Proving the identity: \(\displaystyle{\sin{{3}}}{x}+{\sin{{x}}}={2}{\sin{{2}}}{x}{\cos{{x}}}\)

burubukuamaw

burubukuamaw

Answered question

2022-03-31

Proving the identity:
sin3x+sinx=2sin2xcosx

Answer & Explanation

Drake Huang

Drake Huang

Beginner2022-04-01Added 15 answers

sin3x+sinx=2sin2xcosx
LHS=sin(2x+x)+sinx
=sin2xcosx+sinxcos2x+sinx
=2(sinxcosx)cosx+sin(cos2xsin2x)+sinx
=2sinxcos2x+sinxcos2xsin3x+sinx
=cos2x(2sinx+sinx)sin3x+sinx
=(1sin2x)3sinxsin3x+sinx
=4sinx4sin3x
=4(sinxsinx(1cos2x))
=4(sinxcos2x)
=4sinxcos2x
RHS
=2(2sinxcosx)cosx
=4sinxcos2x

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