Question

Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.

Answer 1

First, notice that:

$2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}$

And in the denominator we have:

$1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}$

then, we have on the original expression:

$\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}$

therefore, the identity equals to sinx