Question

The expression Sin²A+Sin²〈A-B〉-2SinA CosB Sin〈A-B〉 = ? *​

Answer 1

Answer:

Sin^2 B = sin^2 A + sin^2 (A - B) - 2sinA.cosB.sin(A - B)

Where;

Sin(A - B) = sinA.cosB - cosA.sinB,

We have;

sin^2 A + sin^2 (A - B) - 2sinA.cosB.sin(A - B)

= sin^2 A + [sinA.cosB - cosA.sinB]^2 - 2sinA.cosB [sinA.cosB - cosA.sinB]

= sin^2 A + sin^2 A.cos^2 B - 2sinA.cosB.cosA.sinB + cos^2 A.sin^2 B - 2sin^2 A.cos^2 B + 2sinA.cosB.cosA.sinB

= sin^2 A + sin^2 A.cos^2 B + cos^2 A.sin^2 B - 2sin^2 A.cos^2 B

= sin^2 A - sin^2 A.cos^2 B + cos^2 A.sin^2 B

= sin^2 A [1 - cos^2 B] + cos^2 A.sin^2 B

Where;

Sin^2 B + cos^2 B = 1

1 - cos^2 B = sin^2 B, we have;

= sin^2 A.sin^2 B + cos^2 A.sin^2 B

= sin^2 B [sin^2 A + cos^2 A]

= sin^2 B [ 1 ]

= sin^2 B.