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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