An alternative proof of the $\sin(2x)$ identity

Uncle Colin recently explained how he would prove the identity $\sin(2x) \equiv 2 \sin(x)\cos(x)$. Naturally, that isn't the only proof.

@traumath pointed me at an especially elegant one involving the unit circle. Suppose we have an isosceles triangle set up like this:

The vertical 'base' of the triangle is $2\sin(\alpha)$ units; the perpendicular 'height' is $\cos(\alpha)$, both easily established from right-angle trigonometry or (you know) the definitions of sine (roughly speaking, how far up the angle takes you on the unit circle) and cosine (how far across). Its area is $\sin(\alpha)\cos(\alpha)$.

Now suppose we rotate the triangle by an angle of $\alpha$ around the origin like so:


The horizontal base of this triangle is one unit, and the perpendicular height is $\sin(2\alpha)$, making the area of the triangle $\frac 12 \sin(2\alpha)$.

We've only rotated it, so its area hasn't changed -- which means $\sin(\alpha)\cos(\alpha) \equiv \frac 12 \sin(2\alpha)$. Double everything and boom! There's your identity.

* Edited 2016-08-25 to link to the previous post.


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.


4 comments on “An alternative proof of the $\sin(2x)$ identity

  • Jacob Rus

    Fun fact: 2 * sin(angle/2), a.k.a. the “chord”, was the original trigonometric measure studied, not the sine. Hipparchus made the first ever trigonometric table listing the chord of angles in 7.5° increments. Later, Indian mathematicians made finer tables. The word “sine” comes from the Indian word for half a chord (they called the chord “bowstring”), and in general people drew pictures showing the chord vertically, like your first diagram.

    See andā,_koti-jyā_and_utkrama-jyā

    • Colin

      If there was a comment of the day award, this would be a shoo-in :o) Thanks!

  • Robert GT

    I spent a while looking for a similarly nice proof of cos(2x)=cos^2(x)-sin^2(x).

    Turns out there’s a nice one. Starting from the second (rotated) orientation. It uses lengths rather than areas.
    1. Drop a perpendicular to define cos(2x)
    2. Draw the bisector to the midpoint of the chord; call the point of intersection P.
    3. Drop a perpendicular from P. Because of similar triangles, the length along the x-axis from the origin is cos^2(x).
    4. Again by similar triangles, the distance from P to the first perpendicular we dropped is sin^2(x).

    Thus, cos(2x)=cos^2(x)-sin^2(x).

    • Colin

      Thanks, Robert!

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