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

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.

• The formula for $\cos(2\theta)$ is wrong. It should have a minus sign. For the proof you also need $\tan(\theta )= \frac{\sin(\theta)}{\cos(\theta)}$ Feb 16, 2022 at 15:55
• Also you need $\cos^2(\theta) + \sin^2(\theta)=1$ Feb 16, 2022 at 16:02
• Use $\sin2\theta=\frac{\sin2\theta}1$. Then use the hint above to write $1$. Feb 16, 2022 at 16:43
– ACB
Feb 16, 2022 at 17:00

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 $$\cos(2\theta).$$

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

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

• Nice and concise Feb 16, 2022 at 17:15
• @Golden_Ratio: yep, but in fact easier to establish backward.
– user1020730
Feb 16, 2022 at 17:17