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.
