Question

could someone expand tan5theta

Answer 1

\[\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}\]
\begin{eqnarray*}
\tan{3x}&=&\tan{2x+x} \\
&=&\frac{\tan{2x}+\tan{x}}{1-\tan{2x}\tan{x}} \\
&=&\frac{\frac{2\tan{x}}{1-\tan^2{x}}+\tan{x}}{1-\frac{2\tan{x}}{1-\tan^2{x}}\tan{x}} \\
&=&\frac{\frac{2\tan{x}+\tan{x}-tan^3{x}}{1-\tan^2{x}}}{\frac{1-\tan^2{x}-2\tan^2{x}}{1-\tan^2{x}}} \\
\tan{3x}&=&\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}
\end{eqnarray*}
\begin{eqnarray*}
\tan{2x}+\tan{3x}&=&\frac{2\tan{x}}{1-\tan^2{x}}+\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}} \\
&=&\frac{2\tan{x}(1-3\tan^2{x})+(1-\tan^2{x})(3\tan{x}-\tan^3{x}}{(1-\tan^2{x})(1-3\tan^2{x})} \\
&=&\frac{2\tan{x}-6\tan^3{x}+3\tan{x}-\tan^3{x}-3\tan^3{x}+\tan^5{x}}{(1-\tan^2{x})(1-3\tan^2{x})} \\
\tan{2x}+\tan{3x}&=&\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{(1-\tan^2{x})(1-3\tan^2{x})}
\end{eqnarray*}
\begin{eqnarray*}
\tan{2x}\tan{3x}&=&\Big(\frac{2\tan{x}}{1-\tan^2{x}}\Big)\Big(\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}\Big) \\
&=&\frac{2\tan{x}(3\tan{x}-\tan^3{x})}{(1-\tan^2{x})(1-3\tan^2{x})} \\
&=&\frac{2\tan^2{x}(3-\tan^2{x})}{(1-\tan^2{x})(1-3\tan^2{x})}
\end{eqnarray*}
\begin{eqnarray*}
\tan{5x}&=&\tan{2x+3x} \\
&=&\frac{\tan{2x}+\tan{3x}}{1-\tan{2x}\tan{3x}} \\
&=&\frac{\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{(1-\tan^2{x})(1-3\tan^2{x})}}{1-\frac{2\tan^2{x}(3-\tan^2{x})}{(1-\tan^2{x})(1-3\tan^2{x})}} \\
&=&\frac{\tan{2x}+\tan{3x}}{1-\tan{2x}\tan{3x}} \\
&=&\frac{\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{(1-\tan^2{x})(1-3\tan^2{x})}}{\frac{(1-\tan^2{x})(1-3\tan^2{x})-2\tan^2{x}(3-\tan^2{x})}{(1-\tan^2{x})(1-3\tan^2{x})}} \\
&=&\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{(1-\tan^2{x})(1-3\tan^2{x})-2\tan^2{x}(3-\tan^2{x})} \\
&=&\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{1-3\tan^2{x}-\tan^2{x}+3\tan^4{x}-6\tan^2{x}+2\tan^4{x}} \\
\tan{5x}&=&\frac{5\tan{x}-10\tan^3{x}+\tan^5{x}}{1-10\tan^2{x}+5\tan^4{x}}
\end{eqnarray*}

Answer 2

\[e^{i5x}=(e^{ix})^5\]\[
\cos{5x}+i\sin{5x}=(\cos{x}+i\sin{x})^5=\]\[
=\cos^5{x}+5i\cos^4{x}\sin{x}+10i^2\cos^3{x}\sin^2{x}+10i^3\cos^2{x}\sin^3{x}+\]\[+5i^4\cos{x}\sin^4{x}+i^5\sin^5{x}=\]\[=\cos^5{x}-10\cos^3{x}\sin^2{x}+5\cos{x}\sin^4{x}+\]\[+i\left(5\cos^4{x}\sin{x}-10\cos^2{x}\sin^3{x}+\sin^5{x}\right)\]
\[\tan{5x}=\frac{\sin{5x}}{\cos{5x}}=\frac{\sin^5{x}-10\cos^2{x}\sin^3{x}+5\cos^4{x}\sin{x}}{\cos^5{x}-10\cos^3{x}\sin^2{x}+5\cos{x}\sin^4{x}}=\]\[=\frac{\tan^5{x}-10\tan^3{x}+5\tan{x}}{1-10\tan^2{x}+5\tan^4{x}}\]