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


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

[ 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


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

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


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 

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



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