Question

Show that sin²(A+B) - sin ²A-B = sin 2A sin 2B​.

a. sin²(A+B) - sin ²(A-B) = sin 2A sin 2B
b. sin²(A+B) - sin ²(A-B) = sin A sin B
c. sin²(A+B) - sin ²(A-B) = cos A cos B
d. sin²(A+B) - sin ²(A-B) = sin² A - sin² B

Answer 1

To prove that sin²(A+B) - sin²(A-B) = sin 2A sin 2B, we can use the trigonometric identity sin²(A+B) - sin²(A-B) = (sin(A+B))^2 - (sin(A-B))^2 and simplify the expression to obtain 2sin2Acos2B = 2sin2Acos2B. Therefore, option (a) sin²(A+B) - sin²(A-B) = sin 2A sin 2B is the correct answer.

Step-by-step explanation:

To prove that sin²(A+B) - sin²(A-B) = sin 2A sin 2B, we can use the trigonometric identity:

sin²(A+B) - sin²(A-B) = (sin(A+B))^2 - (sin(A-B))^2

Expanding this expression, we get:

(sinAcosB + cosAsinB)^2 - (sinAcosB - cosAsinB)^2

By factoring and simplifying, we obtain:

4sinAcosAsinBcosB = 2sin2Acos2B

Therefore, sin²(A+B) - sin²(A-B) = sin 2A sin 2B. So, option (a) sin²(A+B) - sin²(A-B) = sin 2A sin 2B is the correct answer.