Double angle formulae for $\sin2\theta$ and $\cos2\theta$ in tangent form? [closed]

Question

I am in need of some help.

I have these from textbooks:

• $\sin2\theta = 2\sin\theta \cos\theta$
• $\cos2\theta = \cos^2\theta - \sin^2\theta$
• $\tan2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta} $

I don't remember where, but I then found these tangent representations:

• $\sin2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta}$
• $\cos2\theta = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}$

These tangent representations are then used by substituting $t$ for $\tan^2\theta$ and thus resulting in the equations used in Weierstrass substitution, but I need to either show my work to get to said tangent representations or find and cite a proof. I am struggling in doing both so any help is appreciated. Thank you.

Answer 1

Using $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$,

$\frac{2\tan(\theta)}{1+\tan^2(\theta)}=\frac{2\tan(\theta)}{1}\cdot\frac{1}{1+\tan^2(\theta)}=\frac{2\sin(\theta)}{\cos(\theta)}\cdot\frac{1}{1+\tan^2(\theta)}$

Now

$1+\tan^2(\theta)=1+\frac{\sin^2(\theta)}{\cos^2(\theta)}=\frac{\cos^2(\theta)+\sin^2(\theta)}{\cos^2(\theta)}=\frac{1}{\cos^2(\theta)}$

Replacing this value in the first equation

$\frac{2\tan(\theta)}{1+\tan^2(\theta)} =\frac{2\sin(\theta)}{\cos(\theta)}\cdot\frac{1}{1+\tan^2(\theta)}=\frac{2\sin(\theta)}{\cos(\theta)}\cdot\frac{\cos^2(\theta)}{1}=2\sin(\theta)\cos(\theta)=\sin(2\theta)$

Go through a similar process for $\cos(2\theta).$

Answer 2

$$\sin2\theta=2\sin\theta\cos\theta=\frac{2\sin\theta\cos\theta}{\cos^2\theta+\sin^2\theta}=\frac{2\tan\theta}{1+\tan^2\theta}$$

$$\cos2\theta=\cos^2\theta-\sin^2\theta=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$$