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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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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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 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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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

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