Question

how to prove sina sin(a+2b)-sinb sin(b+2a)=sin(a-b)sin(a+b)

Answer 1

HELLO DEAR,


sina sin(a + 2b) - sinbsin(b + 2a)

⇒ 2 [ sina sin(a + 2b) - sinbsin(b + 2a) ] /2
∴ [ multiply and divide by "2" ]

⇒[ cos(a + 2b - a) - cos(a + 2b + a) - {cos(b + 2a - b) - cos(b + 2a + b) ] /2
∴ [ 2sinAcosB = cos(A - B) - cos(A + B) ]

⇒[ cos(2b) - cos(2a + 2b) - cos(2a) + cos(2a + 2b) ] / 2

⇒[ cos2b - cos2a ] /2
∴ [ cosA - cosB = -2sin(A + B)/2 * sin(A - B)/2 ]

⇒[ -2sin(2a + 2b)/2 * sin(2b - 2a)/2 ] /2

⇒[ -2sin(a + b) * (-)sin(a - b) ] /2

⇒sin(a + b) * sin(a - b)

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Thanks for the question!

It is definitely a very interesting question to solve and do some brainstorming.

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Given,
sina sin(a + 2b) - sinb sin(b + 2a)

= 2 {sina sin(a + 2b) - sinbsin(b + 2a) } /2
= cos(a + 2b - a) - cos(a + 2b + a) - {cos(b + 2a - b) - cos(b + 2a + b) }/2

Since, [ 2sinAcosB = cos(A - B) - cos(A + B) ]

= {cos(2b) - cos(2a + 2b) - cos(2a) + cos(2a + 2b) } / 2

= ( cos2b - cos2a ) /2

Also, we know { cosA - cosB = -2sin(A + B)/2 * sin(A - B)/2 }

= { -2sin(2a + 2b)/2 * sin(2b - 2a)/2 }/2

={ -2sin(a + b) * (-)sin(a - b)} /2

= sin(a + b) * sin(a - b)

Hence, Proved

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Hope it helps and solves your query!!