1. ## T-Substitution [tan(0.5x)]

Can someone show me how using the $t=\tan \left(\frac12 x\right)$ substitution, you can establish:
$\sin x = \frac{2t}{1+t^2}$

$\cos x = \frac{1-t^2}{1+t^2}$

It's memorised in my mind (the subsitution) but I don't know the method which bought these substituion in the first place. Thanks in advance.

2. $\sin x = \frac{2t}{1+t^2} = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}} = \frac{2 \tan \frac{x}{2}}{\sec^2 \frac{x}{2}} = 2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x$

$\cos x = \frac{1-t^2}{1+t^2} = \frac{1- \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}} = \frac{\frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}}{\sec^2 \frac{x}{2}} = \cos^2 \frac{x}{2}-\sin^2 \frac{x}{2} = \cos x$

3. I know how to go from the substitution to prove it but how would I go about finding the substitution for $\sin x$ and $\cos x$ if I was just given $t=\tan \left(\frac12 x\right)$

If I'm just given $t=\tan \left(\frac12 x\right)$ and told to find what $\sin x$ and $\cos x$ would be (in terms of $t$), how would I find it's substitution?

4. You can see a basic way to find them here.

5. Refer to the attached image.

Bobak

Edit: sorry about my funny looking multiplication sign.

6. Originally Posted by bobak
Refer to the attached image.

Bobak
You wrote " $2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \frac{t}{\sqrt{t^2+1}} \color{red}+\color{black}\frac{1}{\sqrt{t^2+1}}$". I think that's a typo.