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Prove the identity: (cosx-sinx)^2 = 1-sin2x

User Anouar
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1 Answer

6 votes
6 votes

Given the identity:


(\cos x-\sin x)^2=1-\sin 2x

The proof of the identity will be as follows:

We will begin from the left-hand side by expanding the square


LHS=(\cos x-\sin x)^2=\cos ^2x-2\sin x\cos x+\sin ^2x

Combine the first and the last terms, which will be the Pythagorean identity


\begin{gathered} =(\sin ^2x+\cos ^2x)-2\sin x\cos x \\ =1-2\sin x\cos x \end{gathered}

Note, note the last term (2 sin x cos x) = sin 2x

So,


\begin{gathered} =1-\sin 2x=R\mathrm{}H\mathrm{}S \\ \end{gathered}

User Pritesh Patel
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