Prove that sin^2(x) = 1/2 – 1/2 cos(2x)

The Mathematician — Thu, 29 Dec 2022 13:00:32 +0000

Show that

\sin^2(x)

is equal to

\frac{1}{2} - \frac{1}{2}\cos(2x)

.

Proof. Recall that we have the next addition formula for Cosine:

\begin{align*} \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B). \end{align*}

Now take

A = B = x

. Then we get:

\begin{align*} \cos(x + x) = \cos(x)\cos(x) - \sin(x)\sin(x) &\iff \cos(2x) = \cos^2(x) - \sin^2(x). \end{align*}

We have seen here that

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

. So we get:

\begin{align*} \cos(2x) = 1 - \sin^2(x) - \sin^2(x) &\iff \cos(2x) = 1 - 2\sin^2(x) \\ &\iff 2\sin^2(x) = 1 - \cos(2x) \\ &\iff \sin^2(x) = \frac{1}{2} - \frac{1}{2}\cos(2x) \end{align*}

Therefore,

\sin^2(x)

is equal to

\frac{1}{2} - \frac{1}{2}\cos(2x)

sin^2(x) = 1/2 - 1/2 cos(2x) Archives - Epsilonify