*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I find it easier to remember trigonometric identities if I can ‘see’ how they fit together. I’m expected to know that $\sin(2x) \equiv 2\sin(x)\cos(x)$, but haven’t been able to prove it. Any ideas?

— Geometry? Right Angles? How About Medians?

Hi, GRAHAM!

My favourite proof jumps out from a picture like this one:

Draw a right-angled triangle with hypotenuse 1 and an angle $x$; then draw its reflection in the adjacent side (so you have an isosceles triangle with angle $2x$ at the apex), as above.

Now, the sides of the triangle are 1, 1 and $2\sin(x)$, at the base; the angles are $\left(\frac{\pi}{2} – x\right)$, $\left(\frac{\pi}{2} – x\right)$ (both at the base on the right) and $2x$ (at the apex on the left).

Using the sine rule, $\frac{1}{\sin\left(\frac{\pi}{2}-x\right)} = \frac{2\sin(x)}{\sin(2x)}$

Cross-multiply: $\sin(2x) = 2 \sin(x) \sin\left(\frac{\pi}{2}-x\right)$.

Since $\sin\left(\frac{\pi}{2}-x\right) = \cos(x)$, that means $\sin(2x) = 2\sin(x)\cos(x)$. With a tombstone!

— Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Stephen Morris

When I saw the diagram I thought you were going to use the area of the triangle = $\frac{1}{2} Base \times height$.

Using horizontal line as the Base it is $\frac{1}{2} \cos(x) \times 2\sin(x)$, using the sloping line as the base it is

$\frac{1}{2}\times 1 \times \sin(2x)$.

Putting these equal gives the result.

* Edited to add in LaTeX

## Mark Dawes

I really like the diagram, but used it in another way.

I started in the same way as Stephen. Using the horizontal line as the base we have two triangles that each have area $\frac{1}{2} \cos(x)\sin(x)$.

Using $\frac{1}{2} a b \sin(C)$ on the big triangle we get an area of $\frac{1}{2} \sin(2x)$.

So $\frac{1}{2} \sin(2x) = \cos(x)\sin(x)$ (because there are two of the smaller triangles), and $\sin(2x) = 2\sin(x)\cos(x)$.

* Edited to fix an error Mark pointed out, and to LaTeX it up.