Prove Prove sin5A−sin A+sin2A=sin2A(2cos3A+1)

Kye 2021-01-13 Answered

Prove Prove sin5AsinA+sin2A=sin2A(2cos3A+1)

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Expert Answer

Obiajulu
Answered 2021-01-14 Author has 98 answers

Notice that if you distribute the sin2A on the right side of sin5AsinA+sin2A=sin2A(2cos3A+1) you get sin5AsinA+sin2A=2sin2Acos3A+sin2A. Since both sides have sin2A, this means we need to rewrite sin5AsinA=2sin2Acos3A.
The Sum-To-Product Formula for a difference of sines is sinusinv=2cos(u+v2)sin((uv2).
Therefore, if sinusinv=sin5AsinA,then u=5Aandv=A. Using the Sum-to-Product Formula then gives:
sin5AsinA+sin2A
=2cos5A+A2sin5AA2+sin2A  Use the Sum-to-Product formula
=2cos6A2sin4A2+sin2A  Simplify.
=2cos3Asin2A+sin2A  Simplify.
=sin2A(2cos3A+1)  Factor out sin2A.

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