Answer
$$\cos3x=4\cos^3x-3\cos x$$
Work Step by Step
$$\cos3x$$
The question asks to write $\sin4x$ in terms of a trigonometric function of $x$. We will need both Sum and Difference Identities and Double-Angle Identities to help here.
$$\cos3x=\cos(2x+x)$$
- Sum Identity for cosine: $\cos(A+B)=\cos A\cos B-\sin A\sin B$ with here $A=2x$ and $B=x$
$$\cos3x=\cos2x\cos x-\sin2x\sin x$$
- For $\sin2x$, we apply Double-Angle Identity for sine: $\sin2A=2\sin A\cos A$ with $A=x$.
- For $\cos2x$, we apply Double-Angle Identity for cosine: $\cos2A=\cos^2A-\sin^2A$.
$$\cos3x=(\cos^2x-\sin^2x)\cos x-(2\sin x\cos x)\sin x$$
$$\cos3x=\cos^3x-\sin^2x\cos x-2\sin^2x\cos x$$
$$\cos3x=\cos^3x-3\sin^2x\cos x$$
- Finally, for $\sin^2x$, we apply Pythagorean Identity: $\sin^2x=1-\cos^2x$
$$\cos3x=\cos^3x-3(1-\cos^2x)\cos x$$
$$\cos3x=\cos^3x-3\cos x+3\cos^3x$$
$$\cos3x=4\cos^3x-3\cos x$$
