Mathematical headaches? Problem solved!
Hi, I'm Colin, and I'm here to help you make sense of maths
Mathematical headaches? Problem solved!
Hi, I'm Colin, and I'm here to help you make sense of maths
Written by Colin+ in big in finland, there's more than one way to do it, triangles, trigonometry.
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.
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 https://en.wikipedia.org/wiki/Chord_(geometry) and https://en.wikipedia.org/wiki/Jyā,_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!