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04-04-2019
Mathematics

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Prove this identity using the product-to-sum identity for sine:
[tex]sin^2x=\frac{1-cos(2x)}{2}[/tex]
umm pls

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LammettHash LammettHash

04-04-2019

It follows from

[tex]\cos2x=\cos(x+x)=\cos x\cos x-\sin x\sin x=\cos^2x-\sin^2x=1-2\sin^2x[/tex]

[tex]\implies\cos 2x-1=-2\sin^2x[/tex]

[tex]\implies\sin^2x=\dfrac{1-\cos2x}2[/tex]

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