Convert trigonometric function to irrational fraction

Question

for example $\cos(\frac\pi6)$ is $\frac{\sqrt3}{2}$. How can I convert any other trigonometric function into this type of fraction and preferably without a calculator?

Answer 1

In general, you can't.

Why $\cos(\frac{\pi}{6})$ has an exact closed-form expression

The angle $\frac{\pi}{6}$ (30°), along with its complement $\frac{\pi}{3}$, (60°) is a special case because the 30-60-90 triangle is easily constructed as half of an equilateral triangle. If the hypotenuse has a length of 1, then one of the legs has a length of $\frac{1}{2}$, and the other has a length of $\sqrt{1^2 - (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ by the Pythagorean Theorem. Thus, the familiar SOH-CAH-TOA rule gives:

$$\sin(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) = \frac{1/2}{1} = \frac{1}{2}$$ $$\sin(\frac{\pi}{3}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}$$ $$\tan(\frac{\pi}{6}) = \cot(\frac{\pi}{3}) = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$ $$\tan(\frac{\pi}{3}) = \cot(\frac{\pi}{6}) = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

What other angles can have their trig functions expressed algebraically?

If you know exact $\sin$ and $\cos$ values for two angles $\alpha$ and $\beta$, you can calculate them for the sum and differences of those angles using the identities:

$$\sin(\alpha \pm \beta) = \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta)$$ $$\cos(\alpha \pm \beta) = \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta)$$

Integer multiples of an angle

When $\alpha = \beta = \theta$, these become the double-angle identities:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$

We can also use $\alpha = 2\theta$ and $\beta = \theta$ to derive triple-angle formulas.

$$\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$$ $$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$

And this process can be extended to any $n \in \mathbb{N}$ to find formulas for $\sin(n\theta)$ and $\cos(n\theta)$, which will be an $n$th-degree polynomial in $\cos(\theta)$ and $\sin(\theta)$.

Bisecting an angle

If $\alpha = \beta = \frac{\theta}{2}$, we get an alternative form of the double-angle formulas:

$$\sin(\theta) = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$$ $$\cos(\theta) = \cos^2(\frac{\theta}{2}) - \sin^2(\frac{\theta}{2})$$

From the latter, in combination with the Pythagorean identity $\cos^2(\frac{\theta}{2}) + \sin^2(\frac{\theta}{2}) = 1$, we get a system of quadratic equations with the solutions:

$$\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}$$ $$\cos(\frac{\theta}{2}) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}$$

Where the choice of signs depends on which quadrant the angle is in.

From these identities, we can bisect any angle with a known cosine. For example, starting with $\cos(\frac{\pi}{2}) = 0$, we have

$$\sin(\frac{\pi}{4}) = \sqrt{\frac{1 - 0}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\cos(\frac{\pi}{4}) = \sqrt{\frac{1 + 0}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

You probably already know $\frac{\pi}{4}$ as a "special" angle from the isosceles right triangle. But we can do another iteration to get $\frac{\pi}{8}$.

$$\sin(\frac{\pi}{8}) = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$$ $$\cos(\frac{\pi}{8}) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

For another example, given the $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ from your question, we can find:

$$\sin(\frac{\pi}{12}) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$$ $$\cos(\frac{\pi}{12}) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$$

Trisecting an angle

The triple-angle formulas from earlier can also be written as:

$$\sin(\theta) = 3\sin(\frac{\theta}{3}) - 4\sin^3(\frac{\theta}{3})$$ $$\cos(\theta) = 4\cos^3(\frac{\theta}{3}) - 3\cos(\frac{\theta}{3})$$

which gives us a cubic polynomial equation for each function.

$$\sin(\frac{\theta}{3}) = x: 4x^3 - 3x + \sin(\theta) = 0$$ $$\cos(\frac{\theta}{3}) = x: 4x^3 - 3x - \cos(\theta) = 0$$

Since these are "depressed" cubics (no quadratic term), they can be solved using Cardano's Formula:

$$x^3 + px + q = 0 \implies x = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}$$

For the $\sin$ equation, we have $p = \frac{-3}{4}$ and $q = \frac{\sin(\theta)}{4}$. Plugging these into the formula gives:

$$\sin({\frac{\theta}{3}}) = \sqrt[3]{-\frac{\sin(\theta)}{8} + \sqrt{\frac{\sin^2(\theta)}{64} - \frac{1}{64}}} + \sqrt[3]{-\frac{\sin(\theta)}{8} - \sqrt{\frac{\sin^2(\theta)}{64} - \frac{1}{64}}}$$

For the $\cos$ equation, we have $p = \frac{-3}{4}$ and $q = \frac{\sin(\theta)}{4}$.

$$\cos(\frac{\theta}{3}) = \sqrt[3]{\frac{\cos(\theta)}{8} + \sqrt{\frac{\cos^2{\theta}}{64} - \frac{1}{64}}} + \sqrt[3]{\frac{\cos(\theta)}{8} - \sqrt{\frac{\cos^2{\theta}}{64} - \frac{1}{64}}}$$

Applying the roots separately to numerators and denominators allows these identities to be simplified to:

$$\sin({\frac{\theta}{3}}) = \frac{1}{2} \left( \sqrt[3]{-\sin(\theta) + \sqrt{\sin^2(\theta) - 1}} + \sqrt[3]{-\sin(\theta) - \sqrt{\sin^2(\theta) - 1}} \right)$$ $$\cos(\frac{\theta}{3}) = \frac{1}{2} \left( \sqrt[3]{\cos(\theta) + \sqrt{\cos^2{\theta} - 1}} + \sqrt[3]{\cos(\theta) - \sqrt{\cos^2{\theta} - 1}} \right)$$

Or, using the Pythagorean identity,

$$\sin({\frac{\theta}{3}}) = \frac{1}{2} \left( \sqrt[3]{-\sin(\theta) + i\cos(\theta)} + \sqrt[3]{-\sin(\theta) - i\cos(\theta)} \right)$$ $$\cos(\frac{\theta}{3}) = \frac{1}{2} \left( \sqrt[3]{\cos(\theta) + i \sin(\theta)} + \sqrt[3]{\cos(\theta) - i \sin(\theta)} \right)$$

Yes, there are $i$'s in there, but because we're adding two conjugate terms, they cancel out and give a real number.

Note that since these formulas involve cube roots instead of just square roots, $\sin(\frac{\theta}{3})$ and $\cos(\frac{\theta}{3})$ are not, in general, constructible numbers. Thus, it is not possible in classical geometry, with its focus on compass and straightedge, to trisect an arbitrary angle.

Quartering an angle

Just bisect it twice.

But for higher-denominator fractions

In general, to find $\sin(\frac{\theta}{n})$ and $\cos(\frac{\theta}{n})$, where $n$ is a prime number, you have to solve an $n$th degree polynomial equation. Unfortunately, there is no general formula to solve polynomials of degree 5 or higher. And because of this, it is not possible to have a closed-form expression for trig functions of any arbitrary rational multiple of $\pi$.

There are, however, some special angles that do have closed-form expressions for their trig functions despite the half-angle and third-angle formulas not applying. For example, $\frac{\pi}{5} = 36°$ has this property

$$\sin(\frac{\pi}{5}) = \sqrt{\frac{5 - \sqrt{5}}{8}} = \frac{\sqrt{2(5 - \sqrt{5})}}{4}$$ $$\cos(\frac{\pi}{5}) = \frac{1 + \sqrt{5}}{4}$$

And from this fact, plus the existence of closed-form solutions for bisecting and trisecting an angle, means that any integer number of degrees ($1° = \frac{\pi}{180}$) should have exact expressions for its trig functions, although it may involve awkward multiple layers of radicals.

Answer 2

To find such values, generally one first learns the very popular trigonometric table:

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After that, using the trigonometric identities such as:
$\rightarrow \sin θ = \frac 1{\operatorname{cosec} θ}$ or $\operatorname{cosec} θ = \frac 1{\sin θ}$ $\qquad$ (Reciprocal)
$\rightarrow \sin^2 θ + \cos^2 θ = 1$ $\qquad$ (Pythagorean)
$\rightarrow \tan θ = \frac {\sin θ}{\cos θ}$ $\qquad$ (Ratio)
$\rightarrow \sin (-θ)=-\sin θ$ $\qquad$ (Opposite Angles)
$\rightarrow \sin (90°-θ)=\cos θ$ $\qquad$ (Complementary Angles)
$\rightarrow \sin (180°-θ)=\sin θ$ $\qquad$ (Supplementary Angles)
$\rightarrow \sin(α+β)=\sin(α)\cos(β)+\cos(α)\sin(β)$ $\qquad$ (Sum & Difference)
$\rightarrow \sin 2θ = 2\sinθ \cosθ$ $\qquad$ (Double Angle)
$\rightarrow \sin (\frac θ2) = ±\sqrt {\frac {1 – \cosθ}2}$ $\qquad$ (Half Angle)
$\rightarrow \sin α + \sin β = 2 \sin\frac {α+β}2 \cos\frac {α-β}2$ $\qquad$ (Product-Sum)
$\rightarrow \sin α\sin β = \frac {\cos (α – β) – \cos (α + β)}2$ $\qquad$ (Products)
and similarly for $\cos$ and $\tan$, etc.

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One can find other trigonometric functions' values in following manner:
$$\mbox{$\bf Example$: $\sin(75°)=?$}$$ Using identity: $\sin(α+β)=\sin(α)\cos(β)+\cos(α)\sin(β)$
We have, $\sin(75°)=\sin(30°)\cos(45°)+\cos(30°)\sin(45°)$
Now looking up the table:
$\sin(75°)=\frac 12.\frac 1 {\sqrt{2}} +\frac{\sqrt 3}2.\frac 1 {\sqrt{2}}$
$$\mbox{$\bf Thus,$ $\sin(75°)=\frac{\sqrt 3+1}{2\sqrt 2}$}$$

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Like with everything else, application of this method takes time to get acquainted with and practice is required so that identities may come to mind at once. Also note that this method is good for certain values and not for others.

Answer 3

By Niven's theorem, the only rational multiples of $\pi$ whose sine or cosine is rational are where the sine or cosine is $0$, $\pm 1/2$ or $\pm 1$.

Now note that $\sin^2(\theta) = (1 - \cos(2\theta))/2$. Thus if $\theta$ is a rational multiple of $\pi$ and $\sin(\theta)$ is the square root of a rational number, $\cos(2\theta)$ must be rational and $2\theta$ is a rational multiple of $\pi$, so $\cos(2\theta)$ is $0$, $\pm 1/2$ or $\pm 1$, leading to $\sin^2(\theta) = 1/2$, $1/4$, $3/4$,$0$ or $1$. Thus the only cases where $\theta$ is a rational multiple of $\pi$ and $\sin(\theta)$ is an irrational square root of a rational number are when $\sin(\theta) = \pm 1/\sqrt{2}$ or $\pm\sqrt{3}/2$. Similarly for $\cos$.