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 ((x)/(2))=2\cdot(\pm\sqrt[]{(1-cosx)/(1+\cos x))}`

And in the denominator we have:

`1+\tan ^2((x)/(2))=1+(1-\cos x)/(1+\cos x)=(1+cosx+1-\cos x)/(1+cosx)=(2)/(1+\cos x)`

then, we have on the original expression:

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