what is the value of sin(2A+2B)
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Answered by
0
Answer:
sin(2A+2B)
= sin2A.cos2B +cos2A.sin2B
Answered by
1
Step-by-step explanation:
We know that cos 2A=1 - 2sin^2A
or 2 sin^2A=1-cos 2A
sin^2A+sin^2B=(1/2)[2sin ^2A+sin ^2B]
=(1/2)[1-cos 2A+1-cos 2B]
=(1/2)[2-{cos 2A+cos 2B}]
=(1/2)[2–2cos(A+B).cos(A-B)]
=1-cos (A+B).cos (A-B) ,
Hope it helps..
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