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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