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ID:
Class: 11
Subject: Math
Topic: Trigonometric Functions
Type: Short
Question:
Prove that \(\qquad \cos 4x = 1 - 8 \sin^2 x \cos^2 x \)
Official Solution
Explanation:
We use the following identities:
\begin{align*}
\cos 2A &= 1 - 2 \sin^2 A \\
\sin 2A &= 2 \sin A \cos A
\end{align*}
Now consider the left-hand side:
\begin{align*}
\cos 4x &= \cos(2 \cdot 2x) \\
&= \cos 2(2x) \\
&= 1 - 2 \sin^2(2x) \\
&= 1 - 2 \left( 2 \sin x \cos x \right)^2 \\
&= 1 - 2 \cdot 4 \sin^2 x \cos^2 x \\
&= 1 - 8 \sin^2 x \cos^2 x
\end{align*}
\textbf{Hence, proved.}
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