# Proof 21: Sine and cosine values

We here derive the sine and cosine values of a few chosen angles.

Half ($\pi$) and quarter turns ($\frac \pi 2$, $-\frac \pi 2$):

Cosine and sine are defined[1] as horizontal and vertical distances, respectively, to reach the unit circle. At an angle of 0, there is no vertical part, so $\sin(0)=0$, while the horizontal part carries the full distance of 1 unit, so $\cos(0)=1$. $\quad \blacksquare$

The same is the case at half a turn, $\pi$, with the one difference that the horizontal part now is “backwards”. So, we give it a negative sign and have: $\cos(0)=-1$ and still $\sin(0)=0$. $\quad \blacksquare$

The two values are “flipped” for a quarter turn, $\pi/2$, where we here have no horizontal part. So, $\cos(\frac \pi 2)=0$ and $\sin(\frac \pi 2)=1$. Likewise with a negative quarter turn, $-\pi/2$, with the only difference that the vertical part is downwards, and so we add a negative sign: $\cos(-\frac \pi 2)=0$ and $\sin(-\frac \pi 2)=-1$.  $\quad \blacksquare$

Halves of quarter turns ($\frac \pi 4$, $\frac{3 \pi}4$, $-\frac \pi 4$, $-\frac{3 \pi}4$):

At an angle of $\pi/4$ (or $45^\circ$), the horizontal and vertical distances to the unit circle are exactly equal. So, the sine and cosine values are the same: $\sin(\frac \pi 4)=\cos(\frac \pi 4)$. Let’s set up Pythagoras’ theorem $a^2+b^2=c^2$, where we know that the hypotenuse ($c$) is 1 (since it is a unit circle):

$$\underbrace{\sin\left(\frac\pi 4\right)^2}_a+ \underbrace{\cos\left(\frac\pi 4 \right )^2}_b=1^2\quad\Leftrightarrow\\ \sin\left(\frac\pi 4 \right )^2+ \underbrace{\sin}_{\cos}\left(\frac\pi 4 \right )^2=1 \quad\Leftrightarrow \\ 2 \sin\left(\frac\pi 4 \right )^2=1 \quad\Leftrightarrow \\ \sin\left(\frac\pi 4 \right )^2=\frac 12 \quad\Leftrightarrow \\ \sin\left(\frac\pi 4 \right )=\sqrt{\frac 12}= \frac{ \sqrt 1}{ \sqrt 2}=\frac 1{ \sqrt 2}=\frac{1\cdot \sqrt 2}{ \sqrt 2\cdot \sqrt 2}=\frac{ \sqrt 2}{ \sqrt 2 ^2}=\frac{ \sqrt 2 }2 \\$$

They are equal, so since $\sin(\frac \pi 4)=\frac {\sqrt 2}2$ then also $\cos(\frac \pi 4)=\frac {\sqrt 2}2$.$\quad\blacksquare$

The half-quarters in the other quadrants have the same values, since they also have identical distances sideways and vertically, but with a negative sign added when “backwards” or downwards: $\cos(\frac{3\pi} 4)=-\frac {\sqrt 2}2$ and $\sin(\frac{3 \pi} 4)=\frac {\sqrt 2}2$, $\cos(-\frac{3 \pi} 4)=-\frac {\sqrt 2}2$ and $\sin(-\frac{3 \pi} 4)=-\frac {\sqrt 2}2$ and finally $\cos(-\frac \pi 4)=\frac {\sqrt 2}2$ and $\sin(-\frac \pi 4)=-\frac {\sqrt 2}2$. $\quad\blacksquare$

Thirds of half turns ($\frac \pi 3$, $\frac{2\pi} 3$, $-\frac \pi 3$, $-\frac{2\pi}3$):

To look at angles of $\pi/3$, we can imagine an equilateral triangle (a triangle with 3 equal sides). All angles in this triangle are equal, so each corner has an angle of one third of the total angle sum. Since the total angle sum in any triangle is $180^\circ$ or $\pi$, each corner is thus $\pi/3$. We now have the angle we want to look at.

Let’s now imagine that this triangle has a side length of 2 units. Cut it through down the middle, and you have two identical right-angled triangles. The hypotenuse is 2 units long. And the short cathetus is 1 unit long. In a right-angled triangle, cosine of an angle equals ‘adjacent side over hypotenuse’, so cosine to the angle of $\pi/3$ is:

$$\cos\left(\frac \pi3\right)=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac 12$$

This is a general result. Had the triangle been, say, twice as big, then both side lengths involved here would have been multiplied by a factor of 2, which would have cancelled out.

We can do the same with sine to the angle, which is ‘opposite side over hypotenuse’. We need the opposite cathetus for this. We can find it with Pythagoras’ theorem:

$$\text{adjacent}^2+\text{opposite}^2=\text{hypotenuse}^2 \quad\Leftrightarrow\quad 1^2+\text{opposite}^2=2^2\quad\Leftrightarrow\\ \text{opposite}^2=4-1=3\quad\Leftrightarrow\quad \text{opposite} =\sqrt 3$$

With this other side length, which is the opposite side from the $\pi/3$ angle, we get:

$$\sin\left(\frac \pi 3\right)=\frac{\text{opposite side}}{ \text{hypotenuse} }=\frac{\sqrt 3}2\quad \blacksquare$$

The equivalent angles that span two thirds of the other quadrants ($\frac{2\pi}3$, $-\frac \pi3$, $-\frac{2\pi}3$), have the same sine and cosine values due to symmetry just with added negative signs when “backwards” or downwards: $\cos( \frac{2\pi}3 )=-\frac 12$ and $\sin( \frac{2\pi}3 )=-\frac{\sqrt 3 \pi}2$, $\cos( -\frac{2\pi}3 )=-\frac 12$ and $\sin( -\frac{2\pi}3 )=-\frac{\sqrt 3 \pi}2$ as well as $\cos( -\frac \pi3 )=\frac 12$ and $\sin( -\frac \pi3 )=-\frac{\sqrt 3 \pi}2$.$\quad \blacksquare$

Thirds of quarter turns ($\frac \pi 6$, $\frac{5\pi}6$, $-\frac \pi 6$, $-\frac{5\pi}6$):

Looking from the other angle in the right-angled triangle, which is half of $\pi/3$, so $\pi/6$, since we cut the triangle through its middle, we get the exact same results, just “flipped”. What before was cosine and sine, now is sine and cosine:

$$\cos\left(\frac \pi 6\right)=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{\sqrt 3}2\\ \sin\left(\frac \pi 6\right)=\frac{\text{opposite side}}{ \text{hypotenuse} }=\frac 12 \quad \blacksquare$$

And as before, the equivalent angles that span one third of the other quadrants ($\frac{5\pi}6$, $-\frac \pi6$, $-\frac{5\pi}6$) have these same sine and cosine values due to the symmetry across the circle, just with an added negative sign when the value is “backwards” or downwards: $\cos(\frac{5\pi}6)=-\frac{\sqrt 3}2$ and $\sin(\frac{5\pi}6)=\frac 12$, $\cos(-\frac{5\pi}6)=-\frac{\sqrt 3}2$ and $\sin(-\frac{5\pi}6)=-\frac 12$, and lastly $\cos(-\frac \pi 6)=\frac{\sqrt 3}2$ and $\sin(-\frac \pi 6)=-\frac 12$.$\quad\blacksquare$

References:

1. Sears and Zemansky’s Univesity Physics with Modern Physics’ (book), Hugh D. Young & Roger A. Freedman, Pearson Education, 13th ed., 2012

#### Responses

1. […] Some useful angles are given in degrees and radians along with their sine and cosine values. Derivations of the values can be found in Proof 21. […]