Third trimester

Third trimester of pregnancy:

Pregnant women and their partners are usually curious in knowing the week in which the third trimester starts ,about how much weight gain can be expected in the third trimester, diet for the third trimester, when to take ultrasound and sonogram in the third trimester, pregnancy pains to expect in the third trimester and what could be the due date of pregnancy .

The period from 27th / 28th week (counting from the  LMP last menstrual period ) to the time of giving birth is called the third trimester. This will usually be in the 40th week though it can be one or two weeks later or earlier usually even for normal births.

Some women start to feel back aches while walking due to the increasing weight of the baby / baby bump . Some women would be able to sleep only on one side due to the developing baby bump. Pregnant mothers who still have to go to work in this trimester would find it hard to get clothes / tops /shirts / shoes that fit comfortably especially if it is the summer season . Plus size clothing in some brands might be useful.  It would also be hard to tie shoelaces.
The eyes of the baby start to open around the 28th week.

The pregnant woman will feel more tired ( increase in fatigue ) in the third trimester when compared to the second trimester. Some pregnant women in the third trimester will have to go to the bathroom frequently, especially at night and this will result in decrease of restful sleep and increase the fatigue

In the third trimester , the baby quickly gains weight (around half a pound or around 225 grams per week ) from around the 34th / 35th week.
There might be more vaginal discharge from the vagina during the third trimester.  You also might pee / leak a little urine when you sneeze or cough ( snissing  sneezing and pissing at the same time )

Eat fresh fruit, vegetables,food with proteins, vitamins and fibre  that can give  you around 250 to 300 calories extra than usual ( concept of eating for two persons ).

Some women  in the third trimester develop Pre eclampsia  ( pre - e or PE ) ( PIH pregnancy induced hypertension )/ toxemia with  high blood pressure ( abrupt increase without having a medical history of hypertension ), high protein presence in the urine, and develop swelling ( edema ) in the hands and legs. If left untreated, it may damage the liver / kidneys of the mother and may cause seizure and this stage is also called eclampasia.

Swelling in legs might also be due to fluid retention in some pregnant women without pre - e .

Around the middle of the third trimester is a good time to interview doulas ( non-medical birth companion who gives support and help to pregnant women during pregnancy and during delivery  ) if you are going to use their services.


It is a good idea to pack an emergency hospital bag around the middle of the  third trimester with things you might need at the hospital.



.


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second trimester of pregnancy

Second trimester of pregnancy

Pregnant ladies and their partners are usually interested in knowing the week in which the second trimester starts ,about how long the second trimester is, how many weeks are there in the second trimester, and if the 13th week is part of the second trimester or not .

The period from 13th / 14th week (counting from the  LMP last menstrual period )to 27th /28th week of pregnancy is usually called the second trimester.

The nausea / morning sickness and fatigue felt during the first trimester should ideally taper off in the second trimester, but this may continue for some pregnant women even into the second trimester.
Some women will experience increased constipation, headaches , nasal congestion / nosebleeds during this time. Drinking a lot of water and food with fibre might be helpful for alleviating constipation. Some women might see a thin white discharge called leukorrhea from the vagina during this trimester. Tampons are usually not advisable during pregnancy for the vaginal discharges. Some women might experience bleeding from the gums due to the increased level of hormones during pregnancy. Some pregnant women might experience a bit of backache. It is better to avoid high heeled footwear and wear comfortable footwear and avoid lifting heavy things.The breasts are likely to get bigger. Some women experience a increase of libido in the second trimester.

Some women feel a sharp stabbing pain in the vaginal region near the end of the second trimester. Some women experience a glowing skin due to increased blood flow during the second trimester. Many women experience heartburn ( a burning sensation from the lower throat through to the oesophagus / food pipe / gullet  ) during the second trimester. Some pregnant women experience round ligament pain / growing pain / pelvic pain (a sharp kind of pain in the lower belly / groin region ) in the second trimester due to stretching of the ligaments for the growing baby in the uterus.

The skeleton / bones  of the baby will begin developing around the 14th/ 15th week. Hair will also begin to develop around this time.

The small baby bump might begin to show up on some women during the second trimester. Around  the  twenty week period the woman might feel the baby moving in her (sometimes called quickening in pregnancy ) and are usually felt as flutters.

The baby is said to be capable of hearing sounds from outside from around the 23rd week.

Working women might need to plan their maternity leave during this trimester. It is better to request your friends or relatives for their unwanted baby clothes / other reusable baby-items around this time, because you are going to find out that stuff for your baby will be a very major expense.

 

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first trimester of pregnancy

First trimester of pregnancy

The period of gestation in a human pregnancy is usually divided into  trimesters starting from the first day of the last menstrual period ( LMP ).

The first three months ( usually through the thirteenth / 12th week ) of pregnancy is usually called the first trimester. This is a very important period and the mother is usually adviced to take extra care. It is best to cut down on alcohol, smoking and caffeine intake.

The implantation will usually start off hormonal changes in the mother's body and this will result in the pregnant mothers body goes through a lot of changes in the first trimester. Nausea, more than usual urination frequency ,higher levels of fatigue, increases sense of smell ,increased sensitivity in the breast / nipple tenderness/  , bluish colouring in the vaginal / labia region are usually seen .The organs of the baby start developing in the first trimester.

Weeks 3 to 8 is usually called the stage of the embryo and from week 9 it is usually referred to as fetus / foetus.

A feeling of bloating / flatulence usually shows up in some pregnant women around the 11th week, because estrogen / progesterone hormone needed for maintaining the pregnancy also relaxes other muscles (in the digestive tract ) leading to slower rates of digestion which might lead to the bloated feeling and farting / gas . Eating small amounts more frequent number of times ,ginger , black tea with lime juice  might be beneficial.

Even though the nausea and possible vomiting are sometimes referred to as morning sickness in pregnancy, the nausea and /or vomiting need not be restricted to the mornings alone

Some pregnant women also feel a metallic taste in the mouth due to the hormonal changes. Some women also express aversion / craving for certain types of food.

In the first trimester pregnant women will usually select their ob - gyn or midwife or stick with their usual family physician. They will usually want to take a first ultrasound or sonogram to check the date the pregnancy, check for twins etc, blood work / analysis, urine analysis, and also will start of the pregnant woman on prenatal vitamins.

But there is also a trend among some doctors to not take an ultrasound or sonogram during the first trimester unless it is absolutely medically necessary because the embryo is most vulnerable in the first trimester stage.

Then some pregnant mothers have to  think about telling others about their pregnancy or to wait till the second trimester before announcing their pregnancy.


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Pregnancy testing

 If you are trying to conceive ( TTC ), and if you are using fertility awareness methods ( FAM )  ,you might have a rough estimation of your day of ovulation , so that you might be able to calculate your days past ovulation ( dpo ).

Women who are trying to concieve ( ttc ) will be having a very nervous two week wait ( TWW ) after the estimated day of ovulation hoping that they will miss the next period or AF ( Aunt flo ).

The standard procedure is usually to wait 15 days past ovulation ( dpo ) with missed periods before testing for pregnancy. But impatient women will usually start testing from ( 7 or 8 dpo) seven or eight days past ovulation using home pregnancy tests ( HPT ). The home pregnancy tests usually test the women's urine for presence of ( HCG ) human chorionic  gonadotropin  hormone produced by the placenta during implantation..

But the home pregnancy tests vary among themselves on their sensitivity ( usually 15 mlU / ml  to 25 mlU /ml ). HCG levels usually will double every two to three days in early pregnancy.

It is better to urinate a little bit first before taking the urine sample for testing from around the middle portion of the urine stream. That means it is better not to use the first part of the urine stream for testing. Some of the home pregnancy tests will involve holding a test stick in the urine stream, while some other HPT 's might involve collecting urine in a collection cup and dipping a strip in the container. Some companies do not recommend using the first morning urine ( FMU ) for testing HPT ( home pregnancy tests ) and OPK ( ovulation prediction ) .

The tests will usually have a control line and presence of another line (sometimes faint) will be the indication of pregnancy. The confirmation of pregnancy is usually called big fat positive ( BFP ) . The doctors might further confirm the pregnancy using  urine pregnancy test ( UPT ) or blood tests or ultrasound / sonogram ( sonogram refers to the picture you get from the ultrasound test  )test and / or a physical pelvic exam .

The ultrasound test can also be used to estimate the age of  the foetus  / fetus and to check the number of fetuses .

The doctor will usually ask for your date of  LMP  ( last menstrual period ) in order to estimate the date of delivery ( EDD ) usually by adding 280 days to the date of the first day of the last menstrual period ( LMP ). The usual gestation period is taken as 40 weeks from the first day of LMP.


Premature deliveries are those that occur before 36 weeks of gestation.



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Trying to get pregnant or trying to concieve ( TTC )

Trying to get pregnant or trying to concieve ( TTC )

Baby making or trying to concieve  (TTC) is a very serious and big business. Some charts try to  find the best window of fertility by using data of basal body temperature ( the initial body temperature taken using a sensitive  and accurate thermometer when you wake up in the morning before getting out of bed hopefully after atleast 3 hours of undisturbed sleep )  and information abour the cervical mucus ( CM ) (which usually is more slippery near ovulation). .

( sample chart  example of bbt from babycenter.com  )

( some pictures of cervical mucus at various stages of menstrual cylce from babycenter.com )


Such methods are sometimes referred to as FAM (fertility awareness methods)

During ovulation, hormonal changes might trigger a rise in BBT ( of around 0.5 degree Farenheit ) , but you will need the data from two or three months to identify that pattern (if any) , because BBT might spike due to other reasons as well.This kind of charting is referred to as temping by some women. BBT will usually spike after ovulation while the cervical mucus slipperiness will show up just before ovulation

The human sperm can usually stay alive in the women's body for around 2 to 3 days (with supporting cervical fluids  inside the uterus ), while the human egg has a life of around only 12 to 24 hours after the egg  is released from the ovary ( ovulation ).

The most fertile period for a woman is  usually only  24 hours before ovulation.

Some ladies use various fertility tracking apps and monitors like Ovia and fertility app, Fertility friend app, Glow (which also uses the male partners health data) and an app called eve , kindara (with the thermometer called wink with the facility of auto synchronisation with the app), Period tracker from itunes etc to track their menstrual cycle and try to find out their most fertile interval so as to maximise the chance of getting pregnant .


Most of these apps are marketed as the most accurate , state of the art and most advanced ovulation calculator, tracker of menstrual periods or period tracker.  But the accuracy and effectiveness of such methods is likely to vary from one woman to the next and it is better for the user to learn about the accuracy and effectiveness of such fertility awareness methods / apps from other sources before they use them.

Women trying to get pregnant use such trackers and apps because they want to find the most efficient period in the monthly cycle in which to try to conceive ( TTC ).

Most of these require you to input basal body temperature (BBT) and/or data on cervical mucus daily etc and use the data to give you best period of trying to conceive. Some of the apps asks for cervical position and nature like softness, how open the cervix is , what kind of mood you have etc? . Putting / tracking all this stuff for the apps might increase stress /  lessen the romantic relationship with your spouse.

Some of these give a basic app for free. But for a fee, you will have access to more detailed analysis.

Then there are some OPK's (ovulation prediction kits) which measure the presence of some luteinizing  hormone ( LH ) in your urine and predict the time of your ovulation (time of release of egg/eggs from the ovary) and hence the window of fertility.
There are some digital OPK's as well. Even if the woman has a regular period, the ovulation  might not happen according to a regular timetable within the menstrual cycle.

Some of the women will be on birth control ( BC ). They will have some notion of a timelime of trying to concieve ( TTC ). So they will have to stop the birth control measures, and allow the normal menstrual cycle to kick in.

Then after TTC there is a TWW ( two week wait ) which is a nervous interval of about two weeks from possible ovulation date before you can test for pregnancy using a pregnancy test kit and hope that this might give a BFP ( big fat positive )


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find the derivative of (sinx+cosx)/(sinx-cosx)

ncert cbse 11th limits and derivatives miscellaneous exercise

17.find the derivative of (sinx+cosx) / (sinx-cosx)

using quotient rule for differentiation

if y = (u/v)

dy/dx = [vu' - uv'] / [v^2]

u = sinx + cosx

v = sinx -  cosx

u' = cosx - sinx

v' = cosx + sinx

y = (sinx+cosx)/sinx-cosx)

dy/dx ={(sinx-cosx)(cosx - sinx) -(sinx+cosx)(cosx + sinx)}/{(sinx-cosx)^2}

on simplification

dy/dx = -2{(sinx)^2 + (cosx)^2} / {(sinx-cosx)^2}

using trigonometry formula trigonometry identities

{(sinx)^2 + (cosx)^2} = 1

dy/dx = (-2) / {(sinx-cosx)^2}

18. find the derivative of (secx -1) / (secx+1)

using quotient rule for differentiation

if y = (u/v)

dy/dx = [vu' - uv'] / [v^2]

u = (secx -1)

v = (secx + 1)

u' = secx tanx

v' = secx tanx

y =(secx -1) / (secx+1)

dy/dx = [vu' - uv'] / [v^2]

dy/dx = [(secx +1)(secx tanx) - (secx -1)(secx tanx) ] / [(secx + 1)^2]

on simplification

dy/dx = [2secx tanx] / [(secx + 1)^2]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

17.find the derivative of (sinx+cosx) / (sinx-cosx)

solution

18. find the derivative of (secx -1) / (secx+1)

solution

23.find the derivative of [(x^2)+1][cosx]

solution

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

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find the derivative of [(x^2)+1][cosx]

ncert cbse 11th limits and derivatives miscellaneous exercise

23.find the derivative of [(x^2)+1][cosx]

y=[(x^2)+1][cosx]

using product rule of differentiation

y = uv

dy/dx = uv' +vu'

u =[(x^2)+1]

v=cosx

u ' =2x+0 = 2x

v ' = (-sinx)

dy/dx = uv' +vu'

dy/dx = [(x^2)+1](-sinx)) +[cosx][2x]

dy/dx = -(x^2)sinx -sinx +2x cosx

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

y=[a(x^2)+sinx ] [p+qcosx]

using product rule of differentiation

y = uv

dy/dx = uv' +vu'

u =[a(x^2)+sinx ]

v =[p+qcosx]

u ' = a(2x)+cosx = 2ax+cosx

v ' =0 - qsinx

dy/dx = uv' +vu'

dy/dx = [a(x^2)+sinx ][ - qsinx] + [p+qcosx][ 2ax+cosx]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

23.find the derivative of [(x^2)+1][cosx]

solution

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

find the derivative of (x+secx)(x-tanx)

ncert cbse 11th limits and derivatives miscellaneous exercise

29.

find the derivative of (x+secx)(x-tanx)

y = (x+secx)(x-tanx)

using product formula for differentiation

y = uv

dy/dx = uv' +vu'

u =(x+secx)

v=(x-tanx)

taking derivative

u' = 1 + secxtanx

v' = 1 - [(secx)^2]

dy/dx = uv' +vu'

dy/dx =(x+secx) {1 - [(secx)^2]} +(x-tanx){1 + secxtanx ]

28.

find the derivative of x / (1+tanx)

using quotient formula for differentiation

y = u/v

dy/dx = { vu' - uv' } / {v^2}

u=x

v=(1+tanx)

taking derivative

u' = 1

v' = 0 + [(secx)^2] =[(secx)^2]

dy/dx = { (1+tanx)(1) - x[(secx)^2] } / {(1+tanx)^2}

dy/dx = { (1+tanx) - x[(secx)^2] } / {(1+tanx)^2}

dy/dx = { vu' - uv' } / {v^2}

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies.

find the derivative of sin(x+1) using first principles

ncert cbse 11th limits and derivatives miscellaneous exercise

1
(iii)find the derivative of sin(x+1) using first principles

f(x) = sin(x+1)

f(x+h) = sin[(x+h) +1]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

=2cos{ [(x+h) +1 +x+1 ] /2 } sin{ [(x+h) +1 - (x+1) ] /2}

=2cos{(2x+h+2)/2}sin{h/2}

[f(x+h) - f(x)] / h =[2cos{(2x+h+2)/2}sin{h/2} / h

=cos{(2x+h+2)/2} * [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x) = cos{(2x+0+2)/2} *[1]

f ' (x) = cos{x+1} cancelling off the 2

(iv) find the derivative of cos[x-(pi/8)] using first principles

f(x) =cos[x-(pi/8)]

f(x+h) =cos[x+h-(pi/8)]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

 using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

=-2sin{ {[x+h-(pi/8)]+[x-(pi/8)]}/2 }sin{ {[x+h-(pi/8)] - [x-(pi/8)]}/2 }

= (-2)sin{ {2[x-(pi/8)]+h}/2 } sin(h/2)

[f(x+h) - f(x)] / h =[ -2sin{ {2[x-(pi/8)]+h}/2 } sin(h/2) ] / h

= -sin{ {2[x-(pi/8)]+h}/2 }* [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x)  = - sin[x-(pi/8)]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution




disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies.

find the derivative of (-x) using first principles

ncert cbse 11th chapter 13 limits and derivatives

1

(i)  find the derivative of (-x) using first principles

let f(x) = (-x)

let h be a small increment in x

f(x+h) = -(x+h)

f(x+h) - f(x) = [ -(x+h)] - [-x] = -x-h+x = (-h)

{f(x+h) - f(x)} / {h} = (-h) / {h} = (-1)

taking limit as h --> 0

f ' (x) = (-1)

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

let f(x) = [ (-x)^(-1) ]

f(x) = (-1) / [x]

f(x+h) = (-1) / [x+h]

f(x+h) - f(x) = [(-1) / [x+h] ] - [ (-1) / [x] ]

={(-x+x+h) / x(x+h)}

=h / x(x+h)

{f(x+h) - f(x)} / {h} = 1 / [ x(x+h) ]

taking limit as h --> 0

f ' (x) = 1 / [x(x+0)]

=1/[(x^2)]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution




disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies.

2(sin(pi/6))^2+(cosec(7pi/6))^2 *[cos(pi/3)]^2 = 3/2

ncert  cbse 11th trigonometry exercise 3.3

2. prove that

2(sin(pi/6))^2+(cosec(7pi/6))^2 *[cos(pi/3)]^2 = 3/2

sin(pi/6) = 1/2

cosec(7pi/6) = cosec[pi + (pi/6) ] = -cosec(pi/6) =( -2) [third quadrant]

cos(pi/3) = 1/2

LHS =2(sin(pi/6))^2+(cosec(7pi/6))^2 *[cos(pi/3)]^2

=2*[(1/2)^2] + [(-2)^2 ] *[(1/2)^2]

=[1/2]+[1]

={3/2} =RHS

exercise3.2

8.  evaluate tan(19pi/3)

tan(19pi/3)  = tan[(18pi + pi) / 3 ]

=tan[6pi + (pi/3)] = tan(pi/3) = [sqrt(3)]


10. evaluate cot(-15pi/4)

cot(-x) = -cotx

cot(-15pi/4) = -cot(15pi/4)

= -cot[(16pi -pi)/4]

= -cot[4pi -(pi/4)]

= -cot(pi/4) = (-1)

3.2

8.  evaluate tan(19pi/3)
solution

10. evaluate cot(-15pi/4)
solution

3.3

1. prove that [sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2 = [-1/2]

  solution

2. prove that

2(sin(pi/6))^2+(cosec(7pi/6))^2 *[cos(pi/3)]^2 = 3/2

solution

4. 2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2  = 10

solution

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

7. [ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2
solution

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2
 solution

9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1
 solution

 
10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2 = 10

ncert  cbse 11th trigonometry exercise 3.3

4. 2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2  = 10

using trigonometry formula trigonometry identities

sin(3pi/4)  = sin[pi - (pi/4)] = sin(pi/4) = 1 / [sqrt(2)]

cos(pi/4) =1 / [sqrt(2)]

sec(pi/3) = 2


LHS =

2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2 

= 2  {1 / [sqrt(2)] }^2 +2  {1 / [sqrt(2)] }^2 +2{2}^2

=2(1/2) +2(1/2) +2(4)

= 1 + 1 + 8 =10  =RHS

1. prove that [sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2 = [-1/2]

using trigonometry formula trigonometry identities

sin(pi/6) =1/2

cos(pi/3) = 1/2

tan(pi/4) = 1

LHS =

[sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2

=[1/2]^2 + [1/2]^2  - [1]^2

= [1/4] +[1/4] - 1

= [1/2] -1

=(-1/2) = RHS

3.3

1. prove that [sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2 = [-1/2]
 solution

4. 2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2  = 10

solution

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

7. [ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2
solution

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2
 solution

9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1
 solution

 
10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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[ tan{(pi/4)+x} ] / [ tan{(pi/4 )- x} ] = { [1+tanx] / [1-tanx] }^2

ncert  cbse 11th trigonometry exercise 3.3

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2

using trigonometry formula trigonometry identities

cos(pi+x) = -cosx [third quadrant]
cos(-x) = cosx  [fourth quadrant]

sin(pi-x) = sinx  [second quadrant]
cos((pi/2)+x) = -sinx [second quadrant]

LHS =  [cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ]

={ [-cosx][cosx]  } / {[sinx] [-sinx]}

=[cosx/sinx]^2

= [cotx]^2


7.prove that
[ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2

using trigonometry formula trigonometry identities

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

tan(pi/4) = 1



LHS =[ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]

[tan(A+B)] / [tan(A-B)] form with A= (pi/4)  B =x

={ [tan(pi/4)+tanx] / [1 -tan(pi/4)tanx] } / {[tan(pi/4)-tanx] / [1+tan(pi/4)tanx]}

use tan(pi/4) = 1

={[1+tanx]/[1-tanx]} / {[1-tanx] / [1+tanx]}

= { [1+tanx] / [1-tanx] }^2 =RHS


9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1

using trigonometry formula trigonometry identities


cos[(3pi/2)+x] = sinx [fourth quadrant ]

cos[2pi+x] = cosx [first quadrant]

cot[(3pi/2)-x] = tanx [third quadrant]

cot[2pi+x] =cotx

LHS =cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]}

=sinx cosx {tanx +cotx }

=sinx cosx { [sinx/cosx] +[cosx/sinx] }

=sinx cosx { [(sinx)^2 + (cosx)^2]  / [cosx sinx] }

=sinx cosx { [1]  / [cosx sinx] }

=1 =RHS

3.3

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

7. [ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2
solution

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2
 solution

9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1
 solution

 

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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tan15 degrees and sin 75 degrees

tan15 degrees and sin 75 degrees

11th cbse ncert trigonometry

5.Find the values of sin 75degrees and tan 15degrees

using trigonometry formula trigonometry identities

sin(A+B) =sinAcosB+cosAsinB


choose A=45degrees B=30degrees

sin(75degrees)
sin(45+30) =sin45cos30+cos45sin30 ( angles in degree )

=[1/(sqrt(2)][sqrt(3)/2] +[1/(sqrt(2)][1/2]

= {[sqrt(3) +1} / {2*sqrt(2)}

using trigonometry formula trigonometry identities 

tan(A-B) = [tanA - tanB] / [1 + tanAtanB]

choose A=45degrees B=30degrees

tan 15degrees

tan[45-30] = [tan45 -tan30]  / [1+tan45tan30] ( angles in degree )

=[1-{1/sqrt(3)}] / [1+(1){1/sqrt(3)}]

=[ sqrt(3) -1 ] / [sqrt(3)} + 1] introduce conjugate

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3) + 1] [ sqrt(3) -1 ] }

={ [ sqrt(3) -1 ]^2 } / { [sqrt(3)}^2 - 1] }
 using identities (a-b)^2 and (a+b)(a-b)
={ [sqrt(3)]^2 - 2*sqrt(3) + 1 } / {3 - 1}

={3-2*sqrt(3) +1} / [2 ]

= [ 4-2*sqrt(3) ] / 2

={ 2 * [2- sqrt(3)]} / 2

= [2 - sqrt(3)]

3.3

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

ncert 11th trigonometry, exercise 3.3

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

using trigonometry formula trigonometry identities

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

LHS = cos[(3pi/4)+x] - cos[(3pi/4)-x]

=2cos[2(3pi/4)/2] cos[2x/2]

=2cos[(3pi/4)] cosx

=2{-1 /(-sqrt(2)) } cosx   [cosx is negative in second quadrant]

= (-sqrt(2))sinx

=RHS

10. prove that

sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx

using trigonometry formula trigonometry identities
cosAcosB +sinAsinB = cos(A-B)

A =(n+2)x
B=(n+1)x


LHS =  sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x]

=cos[A-B] FORM

=cos[(n+2)x - (n+1)x ]

=cosx =RHS


6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)

using trigonometry formula trigonometry identities
cosAcosB - sinAsinB = cos(A+B)

A= [(pi/4)-x]
B= [(pi/4)-y]

LHS =  cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y]

= cos[A+B] form

 = cos[[(pi/4)-x]+[(pi/4)-y]]

=cos[ 2(pi/4)-(x+y)]

=cos[ (pi/2)-(x+y)]

=sin[x+y] = RHS  using cos [ (pi/2)-A] =sinA

3.3

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x exercise 3.3 ncert 11th trigonometry

sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x exercise 3.3 ncert 11th trigonometry

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x

using trigonometry formula trigonometry identities

Rearrange and then use
sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

finally use 1+cosA = 2{ [cos(A/2 ]^2 }

LHS = sin2x+2sin4x+sin6x

=sin2x+sin4x +sin4x+sin6x

={sin4x+sin4x}+{sin6x+sin2x}

={2sin(8x/2)cos(0)}+{2sin(8x/2)cos(4x/2)}

=2sin4x+2sin4xcos2x  [use cos0 = 1]

=2sin4x {1+cos2x}

=2sin4x {2 [ cos(2x/2) ]^2 }

=4[(cosx)^2]sin4x

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]

using trigonometry formula trigonometry identities


sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS= cot4x[sin5x+sin3x]

=cot4x[2sin(8x/2)cos(2x/2)]

=[cos4x/sin4x] [2sin4xcosx]

=2cos4xcosx -------------(1)

RHS=cotx[sin5x-sin3x]

=cotx[2cos(8x/2)sin(2x/2)]

=[cosx/sinx][2cos4xsinx]

=2cos4xcosx -------------(2)

therefore LHS = RHS

3.3

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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exercise 3.3 ncert trigonometry 12

exercise 3.3 ncert trigonometry 12

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

finally use 2sinxcosx  = sin2x

Factorise and then apply the identities

LHS = {sin6x + sin4x } {sin6x - sin4x}

={2sin(10x/2)cos(2x/2)}{2cos(10x/2)sin(2x/2)}

={2sin5xcosx}{2cos5xsinx}

={ 2sinxcosx}{2sin5xcos5x} on regrouping

= {sin2x} {sin10x} using sin2x formula

=RHS

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x

using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

finally use 2sinxcosx  = sin2x

LHS =(cos2x)^2  - (cos6x)^2

={cos2x+cos6x} {cos2x-cos6x}

= { -2sin(8x/2)sin(-4x/2)} {2cos(8x/2)cos(-4x/2)}

= {-2sin4xsin(-2x)}{2cos4xcos(-2x)} {using sin(-x)= -sinx and cos(-x)=cosx}

={-2sin4x[-sin2x]}{2cos4xcos2x}

={2sin2xcos2x}{2sin4xcos4x} on regrouping

=sin4xsin8x

=RHS.

 

3.3

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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ncert 11th cbse trigonometry exercise 3.3 question16

ncert 11th cbse trigonometry exercise 3.3 question16


16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x

using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS = [cos9x -cos5x] / [sin17x - sin3x ]

={ -2sin(14x/2) sin(4x/2) } / { 2cos(20x/2)sin(14x/2) }

= {-2sin7xsin2x} / {2cos10x sin7x}

=  (-sin2x) /cos10x

=RHS

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]

using trigonometry formula trigonometry identities

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

LHS = [sinx -siny] / [cosx +cosy]

={2cos[(x+y)/2] sin [(x-y)/2]} /{2cos[(x+y)/2] cos[(x-y)/2]}

={sin [(x-y)/2]} / { cos[(x-y)/2]}

=tan[(x-y)/2]

=RHS




3.3

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution


19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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exercise 3.3 ncert trigonometry question 19

exercise 3.3 ncert trigonometry question 17 and 19

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x

using trigonometry formula trigonometry identities

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]



LHS =[sin5x + sin3x] / [cos5x+cos3x]

=[2sin(8x/2)cos(2x/2)] / [2cos(8x/2)cos(2x/2)]

= [2sin4xcosx]/[2cos4xcosx]

=sin4x/cos4x

=tan4x


19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x

using trigonometry formula trigonometry identities

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

after rearrangement, after interchanging the terms to make the bigger angle first

LHS = [sinx + sin3x] / [cosx+cos3x]

=[sin3x + sinx] / [cos3x+cosx]

=[2sin(4x/2)cos(2x/2)] / [2cos(4x/2)cos(2x/2)]

=[2sin2xcosx] / [2cos2xcosx]

=sin2x/cos2x

=tan2x

=RHS

3.3

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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exercise 3.3 ncert cbse trigonometry 21

exercise 3.3 ncert cbse trigonometry 21

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x

using trigonometry formula trigonometry identities

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

after grouping terms with 4x and 2x

LHS = [(cos4x+cos2x)+cos3x]/[(sin4x+sin2x)+sin3x]

={2cos(6x/2)cos(2x/2)+cos3x} / {2sin(6x/2)cos(2x/2) + sin3x}

={2cos3xcosx+cos3x} / {2sin3xcosx+sin3x}

={cos3x[2cosx+1]} /{sin3x[2cosx+1]}

=cos3x / sin3x

=cot3x = RHS





3.3

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution 


miscellaneous

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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exercise 3.3 ncert trigonometry

exercise 3.3 ncert trigonometry cbse 11th

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2] in numerator

(cosx)^2 - (sinx)^2 = cos2x in denominator


LHS = [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] introducing (-1) to change the
order in both numerator and denominator

=[sin3x - sinx] / [ (cosx)^2 - (sinx)^2 ]

= [2cos(4x/2) sin(2x/2)]  / [cos2x]

= [2cos(2x) sin(x)]  / [cos2x]

= 2 sinx = RHS





23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}

using trigonometry formula trigonometry identities

tan2x = { 2 tanx }  / { 1 - [(tanx)^2]}
Replace x with 2x

LHS = tan4x =  tan{2(2x)}

= { 2 tan2x }  / { 1 - [(tan2x)^2]} again using the same formula

={ 2[{ 2 tanx }  / { 1 - [(tanx)^2]}] }/ { 1 -[{ 2 tanx }  / { 1 - [(tanx)^2]}]^2 }

expand using identity for (a-b)^2 in denominator . . .

= { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}




3.3

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution 


miscellaneous

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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cbse exercise 3.3 trigonometry problems

cbse exercise 3.3 trigonometry problems

22. Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1

using trigonometry formula trigonometry identities

tan(A+B) = [ tanA + tanB ] / [1-tanAtanB]
and then change [1 / tanx] = cotx

the angles used are x , 2x ,3x

So we connect them using  the relation

3x = x + 2x

take tan on both sides

tan3x = tan( x + 2x )

tan3x = [ tanx + tan2x ]  / [ 1 - tanx tan2x ]

rearranging

[ 1 - tanx tan2x ] (tan3x) = [ tanx + tan2x ]

simplifying

tan3x - tanx tan2x tan3x =[ tanx + tan2x ]

tan3x - tan2x - tanx  = tanx tan2x tan3x

divide each term by  tanx tan2x tan3x

{ 1/[tan2x tanx] }  - { 1 / [tan2x tan3x] } -{ 1/ [tan3x tanx]} =1

using [1 / tanx] = cot x

we get,
 cotx cot2x -cot2xcot3x-cot3xcotx = 1


23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}

using trigonometry formula trigonometry identities

tan2x = { 2 tanx }  / { 1 - [(tanx)^2]}
Replace x with 2x

LHS = tan4x =  tan{2(2x)}

= { 2 tan2x }  / { 1 - [(tan2x)^2]} again using the same formula

={ 2[{ 2 tanx }  / { 1 - [(tanx)^2]}] }/ { 1 -[{ 2 tanx }  / { 1 - [(tanx)^2]}]^2 }

expand using identity for (a-b)^2 in denominator . . .

= { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}




3.3

22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution 


miscellaneous

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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cbse 11th trigonometry exercise 3.3

cbse 11th trigonometry exercise 3.3

24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]

using trigonometry formula trigonometry identities


use the formula cos2x = 1-2(sinx)^2 with x replaced by (2x)
and then use
sin2x = 2 sinx cosx


LHS = cos4x = cos[2(2x)]

= 1-2[ sin(2x) ]^2

= 1 - 2[ 2 sinx cosx ]^2

= 1 -8[(sinx)^2][(cosx)^2] = RHS


25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1

using trigonometry formula trigonometry identities

cos3x = 4[ (cosx)^3] - 3[ cosx ] with x replaced by 2x

and then use  cos2x = 2 [(cosx)^2] - 1

cos6x = cos[3(2x)] =  4[ (cos2x)^3] - 3[ cos2x ]

=4{ [ 2 [(cosx)^2] - 1 ] ^3} - 3{ 2 [(cosx)^2] - 1 }

using the identity for (a-b)^3 = (a^3) -3(a^2)b +3 a(b^2) - (b^3)

=4{ 8  [(cosx)^6]  - 12[(cosx)^4] + 6[(cosx)^2] - 1 } -3{ 2 [(cosx)^2] - 1 }

= 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1  on simplification




3.3

24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution 


miscellaneous

1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution




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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work 

miscellaneous trigonometry question 8 cbse ncert 11th mathematics

miscellaneous trigonometry question 8 cbse ncert 11th mathematics

8.Find  sin(x/2) , cos(x/2) and tan(x/2) if tanx = (-4/3)    x is in the second quadrant.

since x is in the second quadrant

(pi/2) < x < pi so that (pi/4) < (x/2) < (pi/2)

which means that (x/2) is in the first quandrant
and sin(x/2) , cos(x/2) and tan(x/2) are all positive.

given  tanx = (-4/3)
draw a right triangle with x as one of the acute angles
and side opposite to x as 4 units and side adjacent to x is 3 units
using pythagoras theorem hypotenuse = sqrt(16 + 9) = sqrt(25) = 5

since x is in the second quadrant  cosx is negative

cosx = adj / hyp

cosx = (-3/5)

using trigonometry formula trigonometry identities

[sin(x/2)]^2 = (1 - cosx) / 2

[cos(x/2)]^2 = (1 + cosx) / 2

 [sin(x/2)]^2 = [1 - (-3/5) ]  / 2 =[8/5] / 2 = 4/5

[cos(x/2)]^2 = [1 + (-3/5) ]  / 2 =[2/5] / 2 =1/5


take square root and use the fact that (x/2) is in the first quandrant
and sin(x/2) , cos(x/2) and tan(x/2) are all positive.

sin(x/2) = 2 /sqrt(5)

cos(x/2) = 1/sqrt(5)

tan(x/2) = [sin(x/2)]  / [ cos(x/2) ] = [2 /sqrt(5)] /[1 /sqrt(5)] = 2


1.Prove that 2cos(pi/13)cos(9pi/13)+cos (3pi/13)+cos(5pi/13) = 0
solution 

2. Prove that ( sin3x + sinx ) sinx + (cos3x - cosx) cosx = 0

solution

3. Prove that (cosx +cosy)^2 + ( sinx - siny )^2 = 4 { cos[(x+y)/2] }^2

solution

4. Prove that (cosx - cosy)^2 + ( sinx - siny )^2 = 4 { sin[(x-y)/2] }^2

solution 




5.Show that sinx +sin3x+ sin5x +sin7x = 4cosx cos2x sin 4x

solution

6. Show that [sin7x+sin5x +sin9x+sin3x] / [cos7x+cos5x+cos9x+cos3x] = tan6x

solution


7. Prove that sin3x+sin2x-sinx = 4sin(x)cos(x/2)cos(3x/2)
solution 

8.Find  sin(x/2) , cos(x/2) and tan(x/2) if tanx = (-4/3)    x is in the second quadrant.
solution