Sines and cosines of the sum and difference of obtuse angles

Question

For people who know trig a lot you may know the geometric proof of the sines and cosines of the sum and difference of acute angles But i want proof for obtuse angles: Proof 1 is for acute $\alpha$ and $\beta$, with obtuse $\alpha + \beta$ Proof 2 is for acute $\alpha$, with obtuse $\beta$ and $\alpha + \beta \le 180∘$ I have seen here but it does not have the differences written. Also, the proofs have to be a purely Euclidean Geometric proof of the angle sun and difference expansion for sine and cosine, for obtuse angles

Answer 1

I assume that "purely Euclidean geometric" rules out the use of complex numbers, Taylor series definitions of sine and cosine, methods of linear algebra that allow you to construct multiple Cartesian coordinate systems on one plane and construct rotation matrices to transform the coordinates, and so forth. I even assume you don't want to see an explicit Cartesian coordinate system, which rules out the usual unit-circle definitions of trigonometric functions of angles outside the interval $[0,\frac\pi2].$

The problem is that the formulas that are usually defined via a right triangle, such as $\sin(\alpha) = \text{opposite}/\text{hypotenuse},$ are not adequate even to define the trigonometric functions of other angles, let alone derive sum and addition formulas for them. So first of all you must get your mind unstuck from SOHCAHTOA and decide what makes good definitions of the sine and cosine of an obtuse angle.

Keeping in mind that the trigonometric functions are intended to make calculations easy for a wide variety of geometric shapes in a general way, it should not be surprising that we will define the sine and cosine of obtuse angles so that theorems such as the Sine Rule and the Cosine Rule continue to be true for obtuse triangles using the exact same formulas as for acute triangles, saving us from putting a lot of unnecessary effort into special cases of those theorems. What this implies is that the sine of an obtuse angle is equal to the sine of that angle's supplementary angle (which is acute), whereas the cosine of an obtuse angle is a negative number equal in magnitude to the cosine of the supplementary angle (but obviously opposite in sign).

Notice that if $\alpha$ is obtuse then $\alpha = \theta + \frac\pi2$ for some acute angle $\theta,$ and from the above it follows that $\sin(\alpha) = \cos(\theta) = \cos(\alpha - \frac\pi2)$ and $\cos(\alpha) = -\sin(\theta) = -\sin(\alpha - \frac\pi2).$

Here's another question: what about the difference formulas when the difference is negative -- for example, $\sin(\alpha - \beta)$ when $\beta > \alpha$? You don't even need obtuse angles to run into this problem. But you do need to know that we have decided, in order to make trigonometric formulas as generally useful and as free of special cases as possible, that the sine of a negative angle is equal in magnitude but opposite in sign to the sine of the magnitude of the angle, while the cosine of a negative angle is simply the cosine of the magnitude of the angle.

Naturally, from the right triangle definition as well as the facts above, it follows that for acute angles as well as obtuse angles, $\sin(\alpha) = \cos(\frac\pi2 - \alpha) = \cos(\alpha - \frac\pi2)$ and $\cos(\alpha) = \sin(\frac\pi2 - \alpha) = -\sin(\alpha - \frac\pi2).$

Armed with the preceding facts, you can find the $\sin(\alpha - \beta)$ and $\cos(\alpha - \beta)$ for any acute or obtuse angles $\alpha$ and $\beta,$ just by turning the problem into one for which you already have a solution.

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Problem 1. Find $\sin(\alpha - \beta)$ where $\alpha$ is obtuse, $\beta$ is acute, and $\alpha - \beta$ is acute.

Let $\theta = \alpha - \frac\pi2.$ Then $\theta$ is acute, $\alpha = \frac\pi2 + \theta,$ and $$\sin(\alpha - \beta) = \sin\left(\frac\pi2 + \theta - \beta\right) = \sin\left(\frac\pi2 - (\beta - \theta)\right) = \cos(\beta - \theta).$$

If $\beta \geq \theta$ then all three of the angles $\beta,$ $\theta,$ and $\beta - \theta$ are acute and non-negative, so we can rely on the Euclidean proof of the the formula for the cosine of the non-negative acute difference of acute angles as found under How can I understand and prove the "sum and difference formulas" in trigonometry? Therefore \begin{align} \cos(\beta - \theta) &= \cos(\beta)\cos(\theta) + \sin(\beta)\sin(\theta) \\ &= \cos(\beta)\cos\left(\alpha - \frac\pi2\right) + \sin(\beta)\sin\left(\alpha - \frac\pi2\right) \\ &= \cos(\beta)\cos\left(\frac\pi2 - \alpha\right) + \sin(\beta)\left(-\sin\left(\frac\pi2 - \alpha\right)\right) \\ &= \cos(\beta)\sin(\alpha) - \sin(\beta)\cos(\alpha). \end{align} On the other hand, if $\beta < \theta$ then all three of the angles $\beta,$ $\theta,$ and $\beta - \theta$ are acute and non-negative, so we have \begin{align} \cos(\beta - \theta) &= \cos(\theta - \beta) \\ &= \cos(\theta)\cos(\beta) + \sin(\theta)\sin(\beta) \\ &= \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta). \end{align}

In either case, we reach this conclusion: $$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta).$$

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Problem 2. Find $\sin(\alpha - \beta)$ where $\alpha$ is obtuse, $\beta$ is acute, and $\alpha - \beta$ is obtuse.

As in Problem 1, let $\theta = \alpha - \frac\pi2.$ Observing that the the solution to Problem 1 never relied on the premise that $\alpha - \beta$ was acute, we can follow exactly the same procedure with the same results. Conclusion: $$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta).$$

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Problem 3. Find $\sin(\alpha - \beta)$ where $\alpha$ is obtuse, $\beta$ is obtuse, and $\alpha - \beta$ is acute and non-negative.

Let $\theta = \alpha - \frac\pi2$ and $\phi = \beta - \frac\pi2.$ Then $\theta$ and $\phi$ are acute, $\alpha = \frac\pi2 + \theta,$ $\beta = \frac\pi2 + \phi,$ and $\alpha - \beta = \frac\pi2 + \theta - \left(\frac\pi2 + \phi\right) = \theta - \phi.$ The problem statement then implies that $\theta - \phi$ is acute and non-negative. Therefore $$\sin(\alpha - \beta) = \sin(\theta - \phi)$$ and we can rely on the Euclidean proof of the the formula for the sine of the non-negative acute difference of acute angles: \begin{align} \sin(\theta - \phi) &= \sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi) \\ &= \sin\left(\alpha - \frac\pi2\right)\cos\left(\beta - \frac\pi2\right) - \cos\left(\alpha - \frac\pi2\right)\sin\left(\beta - \frac\pi2\right)\\ &= -\cos(\alpha)\sin(\beta) - \sin(\alpha)(-\cos(\beta)). \end{align} Conclusion: $$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta).$$

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Problem 4. Find $\sin(\alpha + \beta)$ where $\alpha$ is acute, $\beta$ is acute, and $\alpha + \beta$ is obtuse.

You have a Euclidean proof under Looking for an alternative proof of the angle difference expansion, but let's see if we can again rely only on the proofs for acute sums of acute angles.

Let $\theta = \frac\pi2 - \alpha.$ Then $\theta$ is acute and non-negative, $\alpha = \frac\pi2 - \theta,$ and the fact that $\alpha + \beta > \frac\pi2$ (given as a premise of the problem statement) implies that $\beta > \frac\pi2 - \alpha = \theta,$ which implies that $\beta - \theta$ is acute and non-negative. Therefore

$$\sin(\alpha + \beta) = \sin(\pi - (\alpha + \beta)) = \sin\left(\pi - \left(\frac\pi2 - \theta + \beta\right)\right) = \sin\left(\frac\pi2 - (\beta - \theta)\right) = \cos(\beta - \theta).$$

We have already found that all three of the angles $\beta,$ $\theta,$ and $\beta - \theta$ are acute and non-negative, so we can rely on the Euclidean proof of the the formula for the cosine of the non-negative acute difference of acute angles, and therefore \begin{align} \cos(\beta - \theta) &= \cos(\beta)\cos(\theta) + \sin(\beta)\sin(\theta) \\ &= \cos(\beta)\cos\left(\frac\pi2 - \alpha\right) + \sin(\beta)\sin\left(\frac\pi2 - \alpha\right) \\ &= \cos(\beta)\sin(\alpha) + \sin(\beta)\cos(\alpha). \end{align}

Conclusion: $$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta).$$

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I could go on like this for quite a lot longer, and would have to do so in order to prove all the remaining formulas involving obtuse angles; we still have not seen the proof for $\sin(\alpha + \beta)$ where one of the angles is obtuse and so is the sum, nor have we seen the cosine formulas proved for any of these cases. But I hope the general idea is clear.

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Personally, I find all these cases tedious. Moreover, what about the sum of two obtuse angles? It is a reflex angle, which we have not yet considered. (The sum of an obtuse angle and an acute angle is also a reflex angle in some cases.) And now that we have to deal with reflex angles, what happens when we add two of them?

An approach that solves all such problems, while still using Euclidean geometry as the underlying basis for proofs, is as follows. First, use Euclidean geometry to develop a Cartesian coordinate system. As a bonus, by constructing Cartesian coordinates this way we can construct any alternative Cartesian coordinate system we want in the same plane. Use Cartesian coordinates and the unit circle to define the sine and cosine of an angle of any magnitude, positive or negative. Use Euclidean geometry (mainly similar right triangles) to establish the equations for conversion of coordinates between a coordinate system $S$ and a coordinate system $S'$ with the same origin but rotated an angle $\theta$ from the system $S.$ Use this mechanism to establish formulas for the coordinates of a point $P'$ that has been obtained by rotating a point $P$ (with given coordinates) through an angle $\theta$ around the origin. Then, keeping in mind that when $\theta < 0,$ "angle $\theta$ counterclockwise" means "angle $\lvert\theta\rvert$ clockwise", take the point $(\cos\alpha,\sin\alpha)$ (which lies at distance $1$ from the origin along a ray at an angle $\alpha$ counterclockwise from the positive $x$-axis), rotate it through an angle $\beta$ counterclockwise, arriving at a point at distance $1$ from the origin on a ray at the angle $\alpha + \beta$ from the positive $x$-axis, and read off the coordinates of that point to get the formulas for $\cos(\alpha + \beta)$ and $\sin(\alpha + \beta).$

This approach requires the development of a lot of mechanics that we don't usually study in so-called "Euclidean geometry" when we study that subject at the age of about fourteen (in a typical U.S. math sequence, at least), but it is an extremely useful set of mechanics for a lot of purposes, and once you have it you can solve all of the angle sum and difference problems for all signs and magnitudes of the angles $\alpha$ and $\beta$ all in one swell foop. I think it is well worth the investment.