Question

Sinx-sin3x+sin5x-sin7x÷cosx-cos3x-cos5x+cos7x=cot2x

Answer 1

Numerator
= sin (x) - sin (3x) + sin (5x) - sin (7x)
= [ sin (x) - sin (3x) ] + [ sin (5x) - sin (7x) ]
= [ 2 cos (2x) sin (-x) ] + [ 2 cos (6x) sin (-x) ]
= -2 cos (2x) sin (x) - 2 cos (6x) sin (x)
= - 2 sin (x) [ cos (2x) + cos (6x) ]
= - 2 sin (x) [ 2 cos (4x) cos (-2x) ]
= - 2 sin (x) [ 2 cos (4x) cos (2x) ]
= - 4 sin (x) cos (4x) cos (2x)

Denominator
= cos (x) - cos (3x) - cos (5x) + cos (7x)
= [ cos (x) - cos (3x) ] - [ cos (5x) - cos (7x) ]
= [ 2 sin (2x) sin (x) ] - [ 2 sin (6x) sin (x) ]
= 2 sin (x) [ sin (2x) - sin (6x) ]
= 2 sin (x) [ 2 cos (4x) sin (-2x) ]
= - 4 sin (x) cos (4x) sin (2x)

Numerator and denominator has a common factor: - 4 sin (x) cos (4x).

Hence, the given expression
= cos (2x) / sin (2x)
= cot (2x)