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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miscellaneous trigonometry question 5 cbse ncert 11th mathematics

miscellaneous trigonometry question 5 cbse ncert 11th mathematics

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

using trigonometry formula trigonometry identities

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

after rearranging

LHS

= (sin 7x + sin x) + (sin5x + sin3x)

= 2 sin(8x/2)cos(6x/2) + 2 sin(8x/2)cos(2x/2)

=2 sin4x cos3x  + 2 sin4x cos x

= 2sin4x [cos 3x + cos x]  taking common factor

= 2sin4x [2 cos(4x/2) cos(2x/2)] using formula cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

=2sin4x [2 cos(2x) cos(x)]

= 4cosx cos2x sin 4x

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

11th cbse trigonometry miscellaneous exercise question 3

11th cbse trigonometry miscellaneous exercise question 3

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

using the trigonometry formulae
cosx + cosy = 2cos[(x+y)/2] cos[(x-y)/2]
sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

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

taking common factor out

= 4{ cos[(x+y)/2] }^2 [{ cos[(x-y)/2] }^2 + { sin[(x-y)/2] }^2 ]

= 4{ cos[(x+y)/2] }^2 [1]

=4{ cos[(x+y)/2] }^2 = RHS.


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

using trigonometry identity

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


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

= 4 { sin[(x-y)/2] }^2 [{ sin[(x+y)/2] }^2 + { cos[(x-y)/2] }^2 ]

= 4 { sin[(x-y)/2] }^2 [ 1 ]

= 4 { sin[(x-y)/2] }^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



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