(sin2(θ))′=sin(2θ)
(\sin ^2(θ))'\:(sin2(θ))′
\mathrm{Apply\:the\:chain\:rule}:\quad 2\sin (θ)(\sin (θ))'\:Applythechainrule:2sin(θ)(sin(θ))′
=2\sin (θ)(\sin (θ))'\:=2sin(θ)(sin(θ))′
(\sin (θ))'\:=\cos (θ)(sin(θ))′=cos(θ)
=2\sin (θ)\cos (θ)=2sin(θ)cos(θ)
\mathrm{Use\:the\:following\:identity}:\quad \:2\cos (x)\sin (x)=\sin (2x)Usethefollowingidentity:2cos(x)sin(x)=sin(2x)
=\sin (2θ)=sin(2θ)
Hope it helps
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Answers & Comments
(sin2(θ))′=sin(2θ)
(\sin ^2(θ))'\:(sin2(θ))′
\mathrm{Apply\:the\:chain\:rule}:\quad 2\sin (θ)(\sin (θ))'\:Applythechainrule:2sin(θ)(sin(θ))′
=2\sin (θ)(\sin (θ))'\:=2sin(θ)(sin(θ))′
(\sin (θ))'\:=\cos (θ)(sin(θ))′=cos(θ)
=2\sin (θ)\cos (θ)=2sin(θ)cos(θ)
\mathrm{Use\:the\:following\:identity}:\quad \:2\cos (x)\sin (x)=\sin (2x)Usethefollowingidentity:2cos(x)sin(x)=sin(2x)
=\sin (2θ)=sin(2θ)
Hope it helps