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3 years ago

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1

Mathematics

1 answer:

miv72 [106K]3 years ago

3 0

Answer:

See below.

Step-by-step explanation:

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1

The 2 halves of this expression are the identities for [sin(A + B)]^2 and

[(cos (A + B)] ^2 respectively , therefore:

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

[(cos (A + B)] ^2

Using the identity sin^2Ф + cos^2Ф = 1 we see that if we put Ф = (A + B) we have

[sin(A + B)]^2 + [(cos (A + B)] ^2 = 1 so the identity

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2 = 1 must be true also.

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((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1​

((sinA.cosB+cosA.sinB)^2)+(cosA.cosB-sinA.sinB)^2=1