# Trig Identities #4 Double Angles

Starting with one of our ‘sum and difference’ identities: $\sin(x\pm y)=\sin(x)\cos(y) \pm \cos(x)\sin(y)$

If we take $\sin(2x) = \sin(x+x)$

then, using the identity we started with: $\sin(2x)=\sin(x)\cos(x) + \cos(x)\sin(x)$ $\sin(2x)= 2\sin(x)\cos(x)$

We can do it all again with cos: $\cos(2x) = \cos(x+x)$ $\cos(2x) = \cos(x)\cos(x) - \sin(x)\sin(x)$ $\cos(2x)=\cos^2(x)-\sin^2(x)$

Now, recall that $\cos^2(x) +\sin^2(x) = 1$ $\cos^2(x) = 1 - \sin^2(x)$

so $\cos(2x) = 1 - \sin^2(x)-\sin^2(x)$ $\cos(2x) = 1 - 2\sin^2(x)$

That’s one identity, now if we go back to the start and manipulate it all a bit differently: $\cos(2x)=\cos^2(x)-\sin^2(x)$ $\cos(2x)=\cos^2(x)-(1-\cos^2(x))$ $\cos(2x)=\cos^2(x) - 1 +\cos^2(x)$ $\cos(2x)=2\cos^2(x) - 1$

Finally, tan is pretty straight forward if you use the sum&difference identity, but we haven’t derived that yet, so here we go: $\tan(x)=\frac{\sin(x)}{\cos(x)}$ $\tan(x + y) = \frac{\sin(x+y)}{\cos(x+y)}$

Using the sin & cos sum/difference identities: $\tan(x + y) = \frac{\sin(x)\cos(y) + \cos(x)\sin(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)}$ $= \frac{\sin(x)\cos(y) + \cos(x)\sin(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)} * \frac{\frac{1}{\cos(x)\cos(y)}} {\frac{1}{\cos(x)\cos(y)}}$ $= \frac{\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)} + \frac{\cos(x)\sin(y)}{\cos(x)\cos(y)} } {\frac{\cos(x)\cos(y)}{\cos(x)\cos(y)} - \frac{\sin(x)\sin(y)}{\cos(x)\cos(y)}}$ $\tan(x + y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$

Phew! Now, if x and y are equal, then just call them both x: $\tan(2x) = \frac{\tan(x)+\tan(x)}{1-\tan(x)\tan(x)}$ $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$