Prove that $\sin x =2t/(1+t^2) $ and $\cos x =(1-t^2)/(1+t^2), t=\tan(x/2)$

Question

Prove that $\sin x =\dfrac{2t}{1+t^2}$ and $\cos x =\dfrac{1-t^2}{1+t^2}$, where $t=\tan\left(\frac{x}{2}\right)$.

Answer 1

Hint : $$1+\tan^2x=\frac{1}{\cos^2x},\quad 2\sin x\cos x=\sin(2x).$$

Answer 2

$$\cos(x)=\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)=\frac{\cos^2\left(\dfrac x2\right)-\sin^2\left(\dfrac x2\right)}{\cos^2\left(\dfrac x2\right)+\sin^2\left(\dfrac x2\right)}=\frac{1-t^2}{1+t^2}.$$

$$\sin(x)=\cos(x)\tan\left(2\frac x2\right)=\frac{1-t^2}{1+t^2}\frac{2t}{1-t^2}.$$

Answer 3

Notice, $$\sin x=\sin 2\left(\frac{x}{2}\right)=\frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$ Now, setting the value of $\tan \frac{x}{2}=t$ as follows $$ \sin x=\frac{2(t)}{1+(t)^2}$$ $$ \implies \color{blue}{\sin x=\frac{2t}{1+t^2}}$$ Again notice, $$\cos x=\cos 2\left(\frac{x}{2}\right)=\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$ Now, setting the value of $\tan \frac{x}{2}=t$ as follows $$ \cos x=\frac{1-(t)^2}{1+(t)^2}$$ $$ \implies \color{blue}{\cos x=\frac{1-t^2}{1+t^2}}$$