Question

Prove that sin9A+sin7A+sin5A+sin3A÷(cos9A+cos7A+cos5A+cos3A)=tan6A

Answer 1

(sin 9x + sin7x + sin 5x + sin 3x)/(cos 9x + cos 7x + cos 5x + cos 3x)

= ({sin 9x + sin 3x} + {sin 7x + sin 5x}) / ({cos 9x + cos 3x} + {cos 7x + cos 5x}) (Sum to Product formula)

=({2 sin [(9x + 3x)/2]*cos[(9x - 3x)/2]}) + ({2 sin[(5x + 7x)/2]*cos[(5x - 7x)]}) /

({2 cos[(9x + 3x)/2]*cos[(9x - 3x)/2]}) + ({2 cos[(5x + 7x)/2]*cos[(5x - 7x)]})

= {(2 sin 6x cos 3x) + (2 sin 6x cos x)} / {(2 cos 6x cos 3x) + (2 cos 6x cos x)}  [cos(-x) = cos x]

= [2 sin 6x (cos 3x + cos x) ] / [ 2 cos 6x ( cos 3x + cos x)]

= 2 sin 6x / 2 cos 6x

= tan 6x

Hence Proved