# Other Exact Trig Values

## A Smaller Piece of π

As we learned very repetitively during the course of study of trigonometry, we are more or less expected to know the trig values of $30$°, $45$°, and $60$°, and consequently their multiples on the unit circle. Later on in our trigonometry study, we were also asked to look at some angle combination formulas and identities. Namely:$$\sin(A \pm B) = \sin(A)\cos(B) \pm \sin(B)\cos(A)$$$$\cos(A \pm B) = \cos(A)\cos(B) \mp \sin(A)\sin(B)$$One of the things you were probably asked to do with these formulas is derive the exact value of the sines and cosines of $15$° and $75$°, but you (almost) certainly were not asked to memorize these results. Namely,$$\sin(15^\circ) = \sin(45^\circ - 30^\circ)$$$$= \sin(45^\circ) \cos(30^\circ) - \sin(30^\circ) \cos(45^\circ)$$$$=\left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \left( \frac{\sqrt{2}}{2} \right)$$$$=\boxed{\frac{\sqrt{6} - \sqrt{2}}{2}}$$$$\cos(15^\circ) = \cos(45^\circ - 30^\circ)$$$$= \cos(45^\circ) \cos(30^\circ) + \sin(45^\circ) \sin(30^\circ)$$$$= \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right)$$$$=\boxed{\frac{\sqrt{6} + \sqrt{2}}{2}}$$$$\sin(75^\circ) = \sin(45^\circ + 30^\circ)$$$$= \sin(45^\circ) \cos(30^\circ) + \sin(30^\circ) \cos(45^\circ)$$$$=\left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{\sqrt{2}}{2} \right)$$$$=\boxed{\frac{\sqrt{6} + \sqrt{2}}{2}}$$$$\cos(75^\circ) = \cos(45^\circ + 30^\circ)$$$$= \cos(30^\circ) \cos(45^\circ) - \sin(30^\circ) \sin(45^\circ)$$$$= \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right)$$$$=\boxed{\frac{\sqrt{6} - \sqrt{2}}{2}}$$So why don't we see these and fill in the unit circle at every interval of $15$°? First of all, because $30$°, $45$°, and $60$° angles have a prominent place in regular geometry figures, such as equilateral triangles and the diagonal cross sections of squares. The same can't be said about $15$° and $75$° angles. Second, due to the fact that the pattern of sine and cosine values as you range from $0$° to $90$° is incredibly kind on the eyes, when you only include $30$°, $45$°, and $60$°. Look closer: You can describe the cosine values from $0$° to $90$° as [root (?) over two] and count from $0$ to $4$.\begin{align} \sin(0^{\circ}) = 0 & = \frac{\sqrt{0}}{2} \\ \sin(30^{\circ}) = \frac{1}{2} & = \frac{\sqrt{1}}{2} \\ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} & = \frac{\sqrt{2}}{2} \\ \sin(60^{\circ}) = \frac{\sqrt{3}}{2} & = \frac{\sqrt{3}}{2} \\ \sin(90^{\circ}) = 1 & = \frac{\sqrt{4}}{2} \end{align}Additionally, the sine values are the same numbers but backward. However if we include $15$° and $75$°, with their awkward root six expressions, it muddles up the picture.

## Unnecessary Roughness

Due to a keen combination of my nerdiness and curiosity, I asked my Pre-Calc teacher if it was possible to find an exact expression for the sine or cosine of $1$° so that you could get an exact expression for any degree measure. Being a bit of a cynical grumpy guy, he curtly told me no. He said all you could do was use half angle formulas and break down from $15$° to $7.5$°, $3.75$°, etc. Of course I took him at his word, but as teachers often do, he was lying his face off.That said, what you're about to see is interesting, but useless in practicality.Without too much difficulty, we can use some quick geometry to derive the sine of $36$°, which leads to the sine of $6$° via the sine difference formula, and then to the sine of $3$° via the half-angle formula.Let's start with an isosceles triangle with base angles of $72$°: Let the bottom base side of the triangle be of length $1$. The congruent legs we will call length $x$. Now, let's draw the angle bisector of the bottom-left base angle: The bisector must be also length $1$ since it is the leg of an isosceles triangle that it created (the blue triangle) with the other leg (the bottom) being $1$. Note also that the green triangle created is isosceles and therefore the other leg of the green triangle is also $1$. Recalling that the entire right leg of the original triangle has its length $x$, this means the base length of the blue triangle must be $x-1$. Next, construct the altitude of the $36$°-$36$°-$108$° triangle, which is also the angle bisector of the $108$° vertex angle, since this $36$°-$36$°-$108$° triangle is also isosceles. Additionally, that altitude bisects the line segment it connects to, once again due to the fact that the triangle is isosceles.  At last, we can use the orange triangle to write an equation for each the sine and cosine of $36$°. $$\sin(36^\circ) = \frac{ \sqrt{1-\big(\frac{x}{2}\big)^2}}{1} = \sqrt{1-\big(\frac{x}{2}\big)^2}$$$$\cos(36^\circ) = \frac{\frac{x}{2}}{1} = \frac{x}{2}$$However, we can find the value of $x$ via proportions. In our figure, the blue triangle is similar to the original big triangle, and therefore the two triangles are proportional. We can solve the following proportion to obtain an exact value of $x$:$$\frac{x}{1} = \frac{1}{x-1}$$$$\longrightarrow x^2 - x - 1 = 0$$$$x = \frac{1 \pm \sqrt{5}}{2}$$However, we can only accept the positive value of this quadratic solution, because $x$ represents a side length in our figure and therefore must be positive.At very long last, we have derived expressions for the exact values of the sine and cosine of $36$°.$$\sin(36^\circ) = \sqrt{1-\big(\frac{\frac{1 + \sqrt{5}}{2}}{2}\big)^2}$$$$=\sqrt{1-\left( \frac{1+\sqrt{5}}{4} \right)^2} = \frac{ \sqrt{16 - (1 + \sqrt{5})^2}}{4}$$$$=\frac{ \sqrt{16 - (6+ 2\sqrt{5})}}{4} = \frac{ \sqrt{10 - 2\sqrt{5}}}{4}$$

$$\cos(36^\circ) = \frac{1}{2} \, \frac{1 + \sqrt{5}}{2}$$$$= \frac{1 + \sqrt{5}}{4}$$Now the fun part: we can use the sine and cosine sum / difference formulas to get sine and cosine values of $6$°. If you're looking for a thrilling good time, you can verify the results:$$\sin(6^\circ) = \sin(36^\circ - 30^\circ)$$$$= \sin(36^\circ)\cos(30^\circ) - \sin(30^\circ) \cos(36^\circ)$$$$= \left( \frac{\sqrt{10-2\sqrt{5}}}{4}\right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \left( \frac{1+\sqrt{5}}{4} \right)$$$$= \frac{\sqrt{30 - 6\sqrt{5}} - (1 + \sqrt{5})}{8}$$$$= \frac{ \sqrt{9 - \sqrt{5} - \sqrt{30 + 6\sqrt{5}}}}{4}$$

$$\cos(6^\circ) = \cos(36^\circ - 30^\circ)$$$$= \cos(36^\circ) \cos(30^\circ) + \sin(36^\circ) \sin(30^\circ)$$$$= \left( \frac{1 + \sqrt{5}}{4}\right) \left( \frac{\sqrt{3}}{2}\right) + \left( \frac{\sqrt{10-s\sqrt{5}}}{4}\right) \left( \frac{1}{2}\right)$$$$= \frac{\sqrt{3}(1 + \sqrt{5}) + \sqrt{10 - 2\sqrt{5}}}{8}$$$$= \frac{\sqrt{7 + \sqrt{5} + \sqrt{30 + 6\sqrt{5}}}}{4}$$Wowee wow. Up for a challenge? Use the half angle and double angle formulas with these results to show that$$\sin(3^\circ) = \frac{\sqrt{8 - \sqrt{3} - \sqrt{15} - \sqrt{10-2\sqrt{5}}}}{4}$$$$\cos(3^\circ) = \frac{\sqrt{8 + \sqrt{3} + \sqrt{15} + \sqrt{10-2\sqrt{5}}}}{4}$$$$\sin(12^\circ) = \frac{\sqrt{8 - 2\sqrt{8 + \sqrt{30+6\sqrt{5}} - \sqrt{6 - 2\sqrt{5}}}}}{4}$$$$\cos(12^\circ) = \frac{\sqrt{8 + 2\sqrt{8 + \sqrt{30+6\sqrt{5}} - \sqrt{6 - 2\sqrt{5}}}}}{4}$$And technically you can get any exact sine or cosine value for any multiple of $3$° by iteratively adding various angles to $3$° using the sum / difference sine and cosine formulas. I won't derive all of those results, but a few years ago someone at ASU did here » if you're curious to see what they all look like (external link).

## One Too Far

For a long time, I thought the exact values of sine and cosine of $3$° was as far as we could go. And as we can see, it's certainly complicated enough. Furthermore, in the real world, folks who actually use trig (engineers, physicists, other smart folks) are doing computer work anyway, making theoretical exact calculations like this purely a matter of satiating the thirst for completeness and exactness.However, if like me you find out you truly can get to the thing you were always curious about, you're going to follow the rabbit hole until you hit its bottom. So let's get wild, child.By using De Moivre's Theorem », we can create an expression for the sine of one-third of an angle. Once we have that monstrous result (and yes, it is genuinely deserving of the name), we can use the result for $\sin(3$°$)$ to obtain an exact-value result that I once believed to be a mathematical unicorn - $\sin(1$°$)$.Let's start with De Moivre's Theorem. It tells us that$$\mathrm{cis}\left(\frac{\theta}{3}\right) = \sqrt{\mathrm{cis}(\theta)}$$$$\longrightarrow ( \cos(\theta / 3) + i\sin(\theta / 3) )^3 = \cos(\theta) + i\sin(\theta)$$$$\longrightarrow \cos(\theta) + i\sin(\theta) =$$$$\cos^3 (\theta / 3) + 3i \cos^2 (\theta / 3) \sin (\theta / 3) - 3\cos(\theta / 3) \sin^2 (\theta / 3) - i \sin^3 (\theta / 3)$$Now, the left hand side real part is $\cos(\theta)$, and it will be equal to the non-imaginary terms on the right side. Similarly, the imaginary part of the left side is $\sin(\theta)$, and it will be equal to the collective imaginary terms on the right side without the $i$. In short,$$\sin(\theta) = 3\sin(\theta / 3) - 4\sin^3 (\theta / 3)$$Ultimately, we have a cubic equation, which is absolutely solvable. It's possible to use the cubic formula » to solve this, but because it is a depressed cubic equation (meaning that there is no quadratic term), it is fairly manageable to solve by brute force.First, let's let $R$ be equal to $\sin(\theta / 3)$, and let $x$ be equal to $\sin(\theta)$. Now we can write our equation as$$R^3 - \frac{3R}{4} + \frac{x}{4} = 0$$Now, let $R = u + v$ such that our equation now reads$$(u + v)^3 - \frac{3(u+v)}{4} + \frac{x}{4}$$If we expand the cubic and also select $u$ and $v$ such that we require $uv = 1/4$, we have$$u^3 + 3u^2 v + 3 u v^2 + v^3 - \frac{3(u+v)}{4} + \frac{x}{4}$$$$u^3 + v^3 + 3uv(u + v) - \frac{3(u+v)}{4} + \frac{x}{4}$$and since we said that whatever we chose for $u$, we would choose $v$ to make sure that $uv = 1/4$, we have$$u^3 + v^3 + \cancel{\frac{3(u+v)}{4}} - \cancel{\frac{3(u+v)}{4}} + \frac{x}{4}$$$$u^3 + v^3 + \frac{x}{4} = 0$$Now, rewrite the result solving for $u^3 + v^3$, and at the same time, note that our requirement that $uv = 1/4$ leads to a requirement for $u^3 v^3$:$$u^3 + v^3 = -\frac{x}{4}$$$$u^3 v^3 = \frac{1}{64}$$Squaring the top equation of these two yields$$u^6 + 2u^3 v^3 + v^6 = \frac{x^2}{16}$$and if we subtract away $4u^3 v^3$ from each side of this we get$$u^6 - 2u^3 v^3 + v^6 = \frac{x^2}{16} - 4u^3 v^3$$On the right hand side, however, let's replace $4u^3 v^3$ with $4 \cdot (1/64)$:$$u^6 - 2u^3 v^3 + v^6 = \frac{x^2}{16} - \frac{4}{64}$$$$u^6 - 2u^3 v^3 + v^6 = \frac{x^2-1}{16}$$Recognizing a perfect square trinomial on the left hand side:$$\left( u^3 - v^3 \right)^2 = \frac{x^2-1}{16}$$Taking the square root of both sides yields$$u^3 - v^3 = \pm \frac{\sqrt{x^2-1}}{4}$$At long last, we have the equations in place to solve for $u$ and $v$, which will allow us to write an expression for $R = u + v = \sin(\theta / 3)$.$$\begin{cases} u^3 + v^3 = -\frac{x}{4} \\ u^3 - v^3 = \pm \frac{\sqrt{x^2-1}}{4} \end{cases}$$Therefore,$$u^3 = -\frac{x}{8} \pm \frac{\sqrt{x^2-1}}{8}$$$$\Rightarrow u = \sqrt{\frac{-x \pm \sqrt{x^2-1}}{8}}$$and$$v^3 = -\frac{x}{8} \mp \frac{\sqrt{x^2-1}}{8}$$$$\Rightarrow v = \sqrt{\frac{-x \mp \sqrt{x^2-1}}{8}}$$$$\therefore R = \sqrt{\frac{-x \pm \sqrt{x^2-1}}{8}} + \sqrt{\frac{-x \mp \sqrt{x^2-1}}{8}}$$Finally we have the needed formula: an expression for $\sin(\theta / 3)$ in terms of $\sin(\theta)$.$$\therefore \sin \left( \frac{\theta}{3} \right) =$$$$\sqrt{\frac{-x \pm \sqrt{x^2-1}}{8}} + \sqrt{\frac{-x \mp \sqrt{x^2-1}}{8}}$$To find the actual exact value of $\sin(1$°$)$, plug the value of $\sin(3$°$)$ into this wonderful formula! That I leave to you. I'm happy enough just knowing the answer exists 😁
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