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


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