Question

Prove it
(cos a - cos 3a)(sin 8a + sin 2a)/ (sin 5a - sin a)(cos 4a - cos 6a) = 1

Answer 1

To prove:

$\frac{(cos a - cos 3a)(sin 8a + sin 2a)}{(sin 5a - sin a)(cos 4a - cos 6a)}$ = 1

Step-by-step explanation:

LHS =  $\frac{(cos a - cos 3a)(sin 8a + sin 2a)}{(sin 5a - sin a)(cos 4a - cos 6a)}$

= $\frac{(2sin\frac{a+3a}{2} . sin\frac{3a-a}{2} )(2sin\frac{8a+2a}{2} . cos\frac{8a-2a}{2})}{(2cos\frac{5a+a}{2} . sin\frac{5a-a}{2})(2sin\frac{4a+6a}{2} . sin\frac{6a-4a}{2} )}$

= $\frac{(2sin\frac{4a}{2} . sin\frac{2a}{2} )(2sin\frac{10a}{2} . cos\frac{6a}{2})}{(2cos\frac{6a}{2} . sin\frac{4a}{2})(2sin\frac{10a}{2} . sin\frac{2a}{2} )}$

= $\frac{4. sin{a} . sin{2a} . sin{5a} . cos{3a}} {4. sin{a} . sin{2a} . sin{5a} . cos{3a}}$

= 1

= RHS

LHS = RHS

Hence proved