Ratios of special angles

Angles worth knowing by heart

A carpenter's set square has angles of $30^\circ$, $60^\circ$ and $90^\circ$. Another has $45^\circ$, $45^\circ$ and $90^\circ$. These angles turn up again and again in buildings, roads and exam papers. Their ratios can be found exactly, with no calculator, from two simple triangles.

45°: half of a square

Cut a square of side $1$ along a diagonal. Each half is a right triangle with two equal legs of $1$ and two angles of $45^\circ$. By Pythagoras, the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$.

For the angle of $45^\circ$, the opposite side is $1$, the adjacent side is $1$ and the hypotenuse is $\sqrt{2}$. So

The size of the square does not matter. With a side of $5\ \text{cm}$, every side is $5$ times longer, and the ratios stay the same.

30° and 60°: half of an equilateral triangle

Take an equilateral triangle of side $2$. Every angle is $60^\circ$. Drop a line from the top to the middle of the base. It cuts the triangle into two right triangles, each with a base of $1$, a hypotenuse of $2$ and a top angle of $30^\circ$. The height is $\sqrt{2^2 - 1^2} = \sqrt{3}$.

For the angle of $60^\circ$ at the base, the opposite side is $\sqrt{3}$ and the adjacent side is $1$:

For the angle of $30^\circ$ at the top, the two legs swap roles. The opposite side is now $1$ and the adjacent side is $\sqrt{3}$:

0° and 90°: the two ends

Keep the hypotenuse at $1$ and make the angle $A$ smaller and smaller. The opposite side shrinks towards $0$, and the adjacent side stretches towards $1$. So we take $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$, and then $\tan 0^\circ = 0$.

Now make $A$ grow towards $90^\circ$. The opposite side grows towards $1$, and the adjacent side shrinks towards $0$. So $\sin 90^\circ = 1$ and $\cos 90^\circ = 0$.

What about $\tan 90^\circ$? It would be $\dfrac{1}{0}$, and we cannot divide by zero. So $\tan 90^\circ$ is not defined. As $A$ nears $90^\circ$, $\tan A$ just keeps growing. In the same way, $\text{cosec}\,0^\circ$, $\cot 0^\circ$ and $\sec 90^\circ$ are not defined.

The table

Here n.d. means not defined. The reciprocal ratios come from this row by row: $\text{cosec}\,A = \dfrac{1}{\sin A}$, $\sec A = \dfrac{1}{\cos A}$ and $\cot A = \dfrac{1}{\tan A}$.

A way to remember. Write $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$. These are the sines of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$. The cosines are the same list read backwards.

Working out an expression

Find the value of $\dfrac{4\cos^2 30^\circ + \tan^2 45^\circ}{\sec^2 60^\circ}$.

1. Write each ratio. $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $\tan 45^\circ = 1$ and $\sec 60^\circ = 2$.
2. Square them. $\cos^2 30^\circ = \dfrac{3}{4}$, $\tan^2 45^\circ = 1$ and $\sec^2 60^\circ = 4$.
3. Put them in. $\dfrac{4 \times \frac{3}{4} + 1}{4} = \dfrac{3 + 1}{4} = 1$.

Finding an angle from a ratio

If $2\sin 3\theta = \sqrt{3}$ and $3\theta$ is acute, then $\sin 3\theta = \dfrac{\sqrt{3}}{2}$. The table gives $3\theta = 60^\circ$, so $\theta = 20^\circ$.

Two equations can fix two angles. Suppose $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \dfrac{1}{\sqrt{3}}$, with $A > B$ and $A + B$ acute. Then $A + B = 60^\circ$ and $A - B = 30^\circ$. Adding gives $2A = 90^\circ$, so $A = 45^\circ$ and $B = 15^\circ$.

Practise with new numbers

Try it in Python

Python's math.sin and math.cos want the angle in radians, so the program turns degrees into radians first.

Common mistakes

• Swapping $\sin 30^\circ$ and $\cos 30^\circ$. $\sin 30^\circ = \dfrac{1}{2}$ is the smaller one, because the side opposite $30^\circ$ is the short leg.
• Forgetting to square. $\sin^2 60^\circ$ means $(\sin 60^\circ)^2 = \dfrac{3}{4}$, not $\dfrac{\sqrt{3}}{2}$.
• Writing $\tan 90^\circ = 0$. It is not defined, since $\cos 90^\circ = 0$.
• Treating $\sin 2A$ as $2\sin A$. $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, but $2\sin 30^\circ = 1$.

Remember these

• $45^\circ$ comes from half a square: sides $1, 1, \sqrt{2}$.
• $30^\circ$ and $60^\circ$ come from half an equilateral triangle: sides $1, \sqrt{3}, 2$.
• As $A$ goes from $0^\circ$ to $90^\circ$, $\sin A$ rises from $0$ to $1$ and $\cos A$ falls from $1$ to $0$.
• $\tan 90^\circ$ is not defined.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is $\tan 60^\circ$?

2. Find $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$.

3. If $\cos 2\theta = 0$ and $2\theta$ is at most $90^\circ$, what is $\theta$?

4. Rohan says $\sin 60^\circ = 2\sin 30^\circ$. Is he right?

5. Why is $\sec 90^\circ$ not defined?