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)


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

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


\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) - 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 + 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)}



Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s