Exact trigonometric values

Why some angles have exact values Extended

A calculator gives $\sin {35}^{°}=0.5735\dots$, a decimal that never ends. But for a few special angles the sine, cosine and tangent can be written exactly, as fractions and square roots. On Paper 2 you have no calculator, so you need these exact values, and you need to be able to use them.

They are not a list to learn blindly. Every one of them comes from two simple triangles that you can sketch in a few seconds: half of a square, and half of an equilateral triangle. If you can draw those two triangles, you can always rebuild the values.

The 45° triangle Extended

Cut a square of side 1 along a diagonal. Each half is a right-angled triangle with two sides of 1, and because it is isosceles its two acute angles are both ${45}^{°}$.

By Pythagoras' theorem, the hypotenuse is $\sqrt{1^2+1^2}=\sqrt{2}$.

From either ${45}^{°}$ angle, the opposite and adjacent sides are both 1, so

• $\sin {45}^{°}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
• $\cos {45}^{°}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
• $\tan {45}^{°}=\frac{1}{1}=1$

$\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are the same number: multiply the top and bottom of $\frac{1}{\sqrt{2}}$ by $\sqrt{2}$ to rationalise the denominator. Either form is exact.

The 30° and 60° triangle Extended

Take an equilateral triangle of side 2. All its angles are ${60}^{°}$. Draw the line of symmetry from the top vertex C to M, the midpoint of the base. It meets the base at a right angle, cuts the base into two pieces of 1, and cuts the top angle into two angles of ${30}^{°}$.

Triangle AMC is right-angled at M, with AC $=2$, AM $=1$, and by Pythagoras' theorem CM $=\sqrt{2^2-1^2}=\sqrt{3}$.

From the ${60}^{°}$ angle at A, the opposite side is $\sqrt{3}$ and the adjacent side is 1:

• $\sin {60}^{°}=\frac{\sqrt{3}}{2}$, $\cos {60}^{°}=\frac{1}{2}$, $\tan {60}^{°}=\frac{\sqrt{3}}{1}=\sqrt{3}$

From the ${30}^{°}$ angle at C, the opposite side is 1 and the adjacent side is $\sqrt{3}$:

• $\sin {30}^{°}=\frac{1}{2}$, $\cos {30}^{°}=\frac{\sqrt{3}}{2}$, $\tan {30}^{°}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

Notice that $\sin {30}^{°}=\cos {60}^{°}$ and $\sin {60}^{°}=\cos {30}^{°}$. The side opposite one acute angle is adjacent to the other, so $\sin x=\cos ({90}^{°}-x)$ always.

0° and 90° Extended

Imagine a right-angled triangle with hypotenuse 1 and an angle $x$ that you make smaller and smaller. The opposite side shrinks towards 0 and the adjacent side grows towards the full length 1. At the limit, $x=0^{°}$:

• $\sin 0^{°}=0$, $\cos 0^{°}=1$, $\tan 0^{°}=\frac{0}{1}=0$

Now open $x$ up towards ${90}^{°}$. The opposite side grows towards 1 and the adjacent side shrinks towards 0:

• $\sin {90}^{°}=1$, $\cos {90}^{°}=0$

$\tan {90}^{°}$ would be $\frac{1}{0}$, which has no value, so tangent is not defined at ${90}^{°}$. That is why the syllabus asks for $\tan x$ only at $0^{°}$, ${30}^{°}$, ${45}^{°}$ and ${60}^{°}$.

The table of exact values Extended

Here is everything together. A pattern helps you check the sine row: it is $\frac{\sqrt{0}}{2}$, $\frac{\sqrt{1}}{2}$, $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, $\frac{\sqrt{4}}{2}$, which simplify to $0$, $\frac{1}{2}$, $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, $1$. The cosine row is the same numbers in reverse order. And each tangent is the sine divided by the cosine, because $\frac{\text{opp}}{\text{hyp}}÷\frac{\text{adj}}{\text{hyp}}=\frac{\text{opp}}{\text{adj}}$.

In the exam, if you are unsure of a value, sketch the triangle in the margin and read it off. It takes ten seconds and it is never wrong.

Sine and cosine of these angles are never more than 1. If you write $\sin {60}^{°}=\sqrt{3}$, the check "sine is at most 1" tells you it is wrong: that is $\tan {60}^{°}$.

Using exact values Extended

Exact values appear in three kinds of question.

• Evaluating an expression. Replace each trig value by its exact value, then simplify with the rules for fractions and surds. Remember that ${\sin }^{2}x$ means $(\sin x{)}^2$, and that $(\sqrt{3}{)}^2=3$.
• Finding a length. Set up the ratio exactly as in any right-angled triangle, then use the exact value instead of a decimal. The answer is exact, often a surd: $12\sin {60}^{°}=12\times \frac{\sqrt{3}}{2}=6\sqrt{3}$.
• Finding an angle. If a ratio comes out as one of the values in the table, read the angle back from the table: $\tan x=\sqrt{3}$ with $x$ acute gives $x={60}^{°}$.

When a length comes out with a root in the denominator, such as $\frac{20}{\sqrt{3}}$, the question may ask you to rationalise it: multiply top and bottom by $\sqrt{3}$ to get $\frac{20\sqrt{3}}{3}$. Simplify any surd fully: $\sqrt{12}=2\sqrt{3}$.