i have been able to do the following

But I'm stuck in simplifying the equation

See my working below

\(\displaystyle \cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x

\)

\(\displaystyle

\begin{align}

\sin3x&=3\cos^2x\sin x-\sin^3x \\[6px]

\cos3x&=\cos^3x-3\cos x\sin^2x \\[12px]

\tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x}

\end{align}

\)

\(\displaystyle

\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}

\)

But I'm stuck in simplifying the equation

See my working below

\(\displaystyle \cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x

\)

\(\displaystyle

\begin{align}

\sin3x&=3\cos^2x\sin x-\sin^3x \\[6px]

\cos3x&=\cos^3x-3\cos x\sin^2x \\[12px]

\tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x}

\end{align}

\)

\(\displaystyle

\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}

\)

Last edited: