The trig summation formulae

The summation formulae for the sine and cosine state that \[\begin{aligned} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha), \: \text{ and } \\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta). \end{aligned}\] While these formulae aren’t used nearly so often as the Pythagorean identity \[\cos^2(\theta) + \sin^2(\theta) = 1,\] they are, nonetheless, of central importance for the development of calculus, since the derivation of the derivative of the sine requires us to expand \(\sin(x+h)\).

Unfortunately, the proofs of the summation formulae aren’t nearly so obvious as the proof the Pythagorean identity.

This wepage scrollytells a proof.

Scroll further to start proof.

We start with a single right triangle containing the angle \(\alpha\).

We then place a second right triangle with angle \(\beta\) whose base lies along the hypotenuse of the first.

We can enclose the whole figure inside a rectangle.

And we can assume that the whole thing is scaled so that the hypotenuse of the second right triangle has length \(1\).

Now, note that the new triangle in the upper right has another angle whose measure is also \(\alpha\) since it, too, is complementary to the angle \(\alpha'\).

Similarly, the new triangle in the upper left has an angle whose measure is \(\alpha+\beta\) since it’s complementary to the angle \(\gamma\).

Now, we can use the basic definitions of the trig functions to express the lengths of the legs of the second triangle as \(\cos(\beta)\) and \(\sin(\beta)\).

And, in fact, we can do the same thing to find the lengths of all the other unlabeled sides.

Finally, simply comparing the total lengths of the left and right sides of the rectangle, we see that \[\begin{aligned}\sin(\alpha+\beta) = & \,\sin(\alpha)\cos(\beta) + \\ &\,\cos(\alpha)\sin(\beta).\end{aligned}\]

Similarly, we can compare the lengths of the top and bottom of the rectangle, to see that \[\begin{aligned}\cos(\alpha+\beta) = & \,\cos(\alpha)\cos(\beta) - \\ & \, \sin(\alpha)\sin(\beta).\end{aligned}\]