Question

prove that sin(a+b)sin(a-b)=sin^2a-sin^2b
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Answer 1

Step-by-step explanation:

sin(A + B)sin(A − B) = sin² A – sin² B -

LHS

= sin(A + B)sin(A - B)

Recall: sin(a - b) = sin a cos ß- cos a sin And sin(a + 3) = sin a cos 3 + cos a sin 3

(sin A cos B+ cos A sin B)

x (sin A cos B - cos A sin B) = sin² A cos² B - cos² A sin² B

Recall: sin² a + cos² a = 1 From above, we can then assume correctly that:

sin² a = 1 cos² a AND -

- cos² a = 1 - sin² a

= sin² A (1 - sin² B) – sin² B(1 – sin² A)

sin² A - sin² A sin² B – sin² B + sin² A sin² B

= sin² A - sin² B

= RHS

Answer 2

To Prove

sin(A+B) * sin(A-B) = sin2 A – sin2 B

Proof

Let us start with LHS given

sin(A+B) * sin(A-B)

We know that formula for sin(A+B) = sin(A+B)=sin(A)cos(B)+cos(A)sin(B)

Also sin(−B)=−sin(B)

cos(−B)=cos(B), so

sin(A−B)=sin(A)cos(B)−cos(A)sin(B)

Therefore by substituting the values in given LHS equation that is sin(A+B)⋅sin(A−B) we get,

=(sinAcosB+cosAsinB)(sinAcosB−cosAsinB)

=(sinAcosB)2−(cosAsinB)2

Now will use the identity (a+b)(a−b)=a2 – b2 in the above equation

=sin2Acos2B−sin2Bcos2A

=sin2A(1−sin2B)−sin2B(1−sin2A)

Now we know that sin2θ+cos2θ=1 (By Pythagoras theorem)

=sin2A−sin2B−sin2A sin2B+sin2Bsin2A

=sin2A−sin2B

sin2A=sun2B