Prove that sin(a+b)*sin(a-b)=sin^2a-sin^2b.

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We have to prove that sin(a + b)*sin(a - b) = (sin a)^2 - (sin b)^2

Use the relation:

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

=> (1/2)*[ cos (a + b - a + b) - cos (a + b + a - b)]

=> (1/2)*[ cos (b + b) -...

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We have to prove that sin(a + b)*sin(a - b) = (sin a)^2 - (sin b)^2

Use the relation:

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

=> (1/2)*[ cos (a + b - a + b) - cos (a + b + a - b)]

=> (1/2)*[ cos (b + b) - cos (a + a)]

=> (1/2)[ cos 2b - cos 2a]

cos 2x = 1 - 2*(sin x)^2

=> (1/2)[1 - 2(sin b)^2 - 1 + 2(sin a)^2]

=> (1/2)[- 2(sin b)^2 + 2(sin a)^2]

=> -(sin b)^2 + (sin a)^2

=> (sin a)^2 - (sin b)^2

This proves : sin(a + b)*sin(a - b) = (sin a)^2 - (sin b)^2

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