Question

sinAsin2a+sin3asin6a-sin4asin5a=0

Answer 1

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Answer 2

2 answers · Mathematics 

 Best Answer

I am trying to explain all the steps. I hope you can understand them-- 

sin A sin 2A + sin 3A sin 6A/ sin A cos 2A + sin 3A cos 6A = tan 5A 

We will use the following formulae in this solution-- 

(1) Sin A . Sin B = (1/2) [ Cos (A-B) - Cos (A+B) ] 
(2) Sin A Cos B = (1/2) [ Sin (A+B) + Sin (A-B) ] 
(3) Sin A + Sin B = 2 Sin (1/2) (A+B) . Cos (1/2)(A-B) 
(4) Sin A - Cos B = 2 Sin (1/2) (A-B) . Cos (1/2)(A+B) 
(5) Sin ( -x ) = - Sin x 
(6) Cos ( - x ) = Cos x 

Solution -- 

Numerator = (1/2) [ 2 SinA Sin 2A + 2 Sin 3A. Sin 6A ] --- Use formula (1) 

=> (1/2) [ Cos (A-2A) - Cos (A+2A) + Cos (3A-6A) - Cos (3A+6A) ] 

=> (1/2) [ Cos A - Cos 3A + Cos 3A - Cos 9A ] 

=> (1/2) [ Cos A - Cos 9A ] .... Again use formula (1) 

=> Sin 5A . Sin 4A 

Denominator 

= Sin A Cos 2A + Sin 3A.Cos 6A 

=> (1/2) [ Sin (A+2A) + Sin (A-2A) + Sin ( 3A + 6A ) + Sin ( 3A - 6A ) ] 

=> (1/2) [ Sin 3A - Sin A + Sin 9A - Sin 3A ] 

=> (1/2) [ Sin 9A - Sin A ] 

=> (1/2) [ 2 Cos 5A . Sin 4A ] 

=> Cos 5A . Sin 4A 

Hence the Left Hand Side of the given identity 

= Numerator / Denominator 

=> Sin 5A . Sin 4A / Cos 5A . Sin 4A 

=> Tan 5A = RHS ………………. QED