# Some Trig

A large portion of the “standard” precalculus curriculum consists of a rather tedious recitation of trigonometric identities. I am largely of the opinion that there are only a couple that one really needs to know—for example, the Pythagorean identity ($$\sin(\theta)^2 + \cos(\theta)^2 = 1$$ for all $$\theta$$), one of the angle addition formulæ (e.g. $$\cos(\theta+\varphi) = \cos(\theta)\cos(\varphi) – \sin(\theta)\sin(\varphi)$$ for all $$\theta,\varphi$$), and the law of cosines (essentially the Pythagorean theorem with a correction term) immediately come to mind as important results. In the last quarter, while teaching precalculus class, I was introduced to a couple of proofs that I had not seen before. I don’t get the impression that either proof is that unusual or original, but I do think that they are both rather slick. I’m putting them up here for future reference.

### Angle Addition Formulæ

Theorem: Given any two angles $$\alpha, \beta$$, we have the identities $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) – \sin(\alpha)\sin(\beta) \quad\text{and}\quad \sin(\alpha + \beta) = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta).$

Proof: It is possible to obtain angle addition formulæ for both sines and cosines with one picture. The argument is as follows: first, we need to angles to add together, say $$\alpha$$ and $$\beta$$. We are going to draw them so that $$\alpha$$, $$\beta$$, and $$\alpha+\beta$$ all live in quadrant I, though the argument can be generalized to other quadrants without too much additional work (exercise: do this).

Note that the point $$B$$ is the point on the unit circle where the ray emanating from the origin with angle $$\alpha+\beta$$ intersects, hence $B = (\cos(\alpha+\beta),\sin(\alpha+\beta));$ we’ll come back to this later. Next, we can construct a line through $$B$$ that is perpendicular to the ray emanating from the origin with angle $$\alpha$$—call the intersection $$C$$.

Similarly, construct a segment perpendicular to the $$x$$-axis through $$C$$, and a segment perpendicular to the $$y$$-axis through $$B$$—these are the points $$D$$ and $$E$$, respectively.

We finish the figure by extending the segments $$EB$$ and $$DC$$ until they intersect at a point $$F$$. Note that we now have a rectangle $$ODFE$$.

The line $$OB$$ is transverse to the parallel lines $$EF$$ and $$OD$$ (since these are opposite sides of a rectangle), thus $$m\angle EBO = m\angle COD = \alpha+\beta$$. Moreover, since $m\angle FCB + m\angle BCO + m\angle OCD = \pi,$ it follows that $$m\angle FCB = \alpha$$. These are labeled in the next figure.

Recall that $$B = (\cos(\alpha+\beta),\sin(\alpha+\beta))$$. This gives us the lengths of the segments $$OE$$ and $$EB$$.

Via some right triangle trig identities on $$\triangle OCB$$ (in particular, the identities given by the mnemonic “SOH CAH TOA”), we have that $$OC = \cos(\beta)$$ and $$BC = \sin(\beta)$$.

Using the same trig identities, this time applied to $$\triangle ODC$$, we have that $$OD = \cos(\alpha)\sin(\beta)$$, and $$CD = \sin(\alpha)\cos(\beta)$$.

Finally, repeating the same argument on $$\triangle BFC$$, we have $$BF = \sin(\alpha)\sin(\beta)$$ and $$CF = \cos(\alpha)\sin(\beta)$$.

But this figure essentially completes the proof. Since opposite sides of a rectangle have the same length, the vertical sides give us $\sin(\alpha+\beta) = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta),$ and the horizontal sides render $\cos(\alpha+\beta) + \sin(\alpha)\sin(\beta) = \cos(\alpha)\cos(\beta).$ Note that, after moving the $$\sin(\alpha)\sin(\beta)$$ term of the second identity to the right-hand side, these are the proposed angle addition formulæ.

### Law of Cosines

Theorem: Suppose that an arbitrary triangle has sides of length $$a$$, $$b$$, and $$c$$, with the angle $$\theta$$ opposite the side $$c$$. Then $c^2 = a^2 + b^2 – 2ab\cos(\theta).$ Note that if $$\theta$$ is a right angle, then $$c$$ is the hypotenuse of a right triangle and $$\cos(\theta) = 0$$, giving the Pythagorean Theorem. Hence we may regard the Law of Cosines as a generalization of the Pythagorean Theorem with some kind of correction term.

Proof: We may assume that the triangle is oriented such that $$\theta$$ is at the origin and one of the two “legs” is along the $$x$$-axis in the Cartesian plane. Labeling the vertices such that vertex $$A$$ is opposite the side with length $$a$$, $$B$$ is opposite $$b$$, and $$C$$ is opposite $$c$$, we obtain the figure shown below.

Since the point $$A$$ is along the $$x$$-axis, the $$y$$-coordinate of $$A$$ is zero, and since the distance from $$A$$ to the origin ($$C$$) is the length of the corresponding leg of the triangle, we have that the $$x$$-coordinate of $$A$$ is $$b$$. That is,$A = (b,0).$ To determine the coordinates of $$B$$, note that the ray from $$C$$ through $$B$$ must intersect the unit circle at some point $$B’$$, as shown below.

From the definitions of sine and cosine on the unit circle, we know that the coordinates of $$B’$$ are given by $$B’ = (\cos(\theta),\sin(\theta))$$. Then, as $$\triangle ABC$$ is similar to $$\triangle AB’C$$, it follows that $B = (a\cos(\theta),a\sin(\theta)).$

On the one hand, we know that $$d(A,B) = c$$ (that is, the distance from $$A$$ to $$B$$ is $$c$$ units). On the other hand, the distance formula tells us that $d(A,B) = \sqrt{(a\cos(\theta)-b)^2 + (a\sin(\theta))^2} = \sqrt{a^2(\cos(\theta)^2 + \sin(\theta)^2) – 2ab\cos(\theta) + b^2}.$ Applying the Pythagorean Identity $$\cos(\theta)^2 + \sin(\theta)^2 = 1$$ and squaring, we obtain $c^2 = d(A,B)^2 = a^2 – 2ab\cos(\theta) + b^2,$ which is the desired result.

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