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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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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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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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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trigonometry miscellaneous question 2 ncert cbse 11th

trigonometry miscellaneous question 2 ncert cbse 11th

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

Using trigonometry formula trigonometry identities
sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]
cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

LHS = {2sin[(3x+x)/2] cos[(3x-x)/2]}sinx +{-2sin[(3x+x)/2] sin[(3x-x)/2]} cosx

=2sin2xcosxsinx - 2sin2xsinxcosx = 0 = RHS


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



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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