Exact trigonometric values

An exact trigonometric value contains no rounding. For special angles, values such as

$$\sin 30^\circ=\frac12, \qquad \cos 45^\circ=\frac{\sqrt2}{2}, \qquad \tan 60^\circ=\sqrt3$$

are exact, whereas $0.7071$ and $1.732$ are approximations. These values should be known and understood, not treated as calculator facts.

Prerequisites

You should be able to:

• use Pythagoras’ theorem and the ratios sine, cosine and tangent from Pythagoras and trigonometry;
• simplify roots and rationalise denominators using surds;
• convert between degrees and radians.

The exact-value table

The five special angles in the first quadrant are:

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
radians	$0$	$\pi/6$	$\pi/4$	$\pi/3$	$\pi/2$
$\sin\theta$	$0$	$1/2$	$\sqrt2/2$	$\sqrt3/2$	$1$
$\cos\theta$	$1$	$\sqrt3/2$	$\sqrt2/2$	$1/2$	$0$
$\tan\theta$	$0$	$1/\sqrt3=\sqrt3/3$	$1$	$\sqrt3$	undefined

Three patterns make the table easier to reconstruct.

1. The sine numerators are $\sqrt0,\sqrt1,\sqrt2,\sqrt3,\sqrt4$, all divided by $2$.
2. The cosine row is the sine row in reverse.
3. Use $\tan\theta=\sin\theta/\cos\theta$ rather than memorising a separate pattern.

The first pattern is a memory aid, not an explanation. The geometry below explains why the values are true.

Deriving the values for $45^\circ$

Take a right-angled isosceles triangle with shorter sides $1$ and $1$. Both acute angles are $45^\circ$. By Pythagoras, its hypotenuse is

$$\sqrt{1^2+1^2}=\sqrt2.$$

Therefore

$$\sin45^\circ=\frac{1}{\sqrt2}=\frac{\sqrt2}{2},$$ $$\cos45^\circ=\frac{1}{\sqrt2}=\frac{\sqrt2}{2},$$

and

$$\tan45^\circ=\frac11=1.$$

The form $1/\sqrt2$ is exact, but $\sqrt2/2$ is the usual rationalised form.

Deriving the values for $30^\circ$ and $60^\circ$

Start with an equilateral triangle of side length $2$. Bisect it from a vertex to the opposite side. This produces two congruent right-angled triangles with hypotenuse $2$, base $1$, and angles $30^\circ$, $60^\circ$ and $90^\circ$.

If the remaining side is $h$, then

$$h^2+1^2=2^2,$$

so

$$h=\sqrt3.$$

The side ratio is therefore

$$1:\sqrt3:2.$$

Using the $30^\circ$ angle,

$$\sin30^\circ=\frac12, \qquad \cos30^\circ=\frac{\sqrt3}{2}, \qquad \tan30^\circ=\frac1{\sqrt3}=\frac{\sqrt3}{3}.$$

Using the $60^\circ$ angle swaps the opposite and adjacent sides:

$$\sin60^\circ=\frac{\sqrt3}{2}, \qquad \cos60^\circ=\frac12, \qquad \tan60^\circ=\sqrt3.$$

Self check 1

Without a calculator, find:

1. $\cos(\pi/3)$
2. $\sin(\pi/4)$
3. $\tan(\pi/6)$
4. $\dfrac{\sin60^\circ}{\cos60^\circ}$

Answers

1. $\cos(\pi/3)=\cos60^\circ=1/2$.
2. $\sin(\pi/4)=\sin45^\circ=\sqrt2/2$.
3. $\tan(\pi/6)=\tan30^\circ=\sqrt3/3$.
4. $\dfrac{\sqrt3/2}{1/2}=\sqrt3$, agreeing with $\tan60^\circ$.

Why $0^\circ$ and $90^\circ$ need the unit circle

A right triangle becomes degenerate at $0^\circ$ or $90^\circ$, so use the unit circle, the circle of radius $1$ centred at the origin. The point reached by turning through an angle $\theta$ has coordinates

$$(\cos\theta,\sin\theta).$$

At $0^\circ$, the point is $(1,0)$, so

$$\cos0^\circ=1, \qquad \sin0^\circ=0, \qquad \tan0^\circ=\frac01=0.$$

At $90^\circ$, the point is $(0,1)$, so

$$\cos90^\circ=0, \qquad \sin90^\circ=1.$$

But

$$\tan90^\circ=\frac{\sin90^\circ}{\cos90^\circ}=\frac10,$$

which is undefined. It is not infinity and it is not zero.

Exact values in every quadrant

The special triangles provide the magnitude of a value. Its sign comes from the coordinates $(x,y)=(\cos\theta,\sin\theta)$ on the unit circle.

Quadrant	Angle range	$\sin\theta$	$\cos\theta$	$\tan\theta$
I	$0^\circ<\theta<90^\circ$	$+$	$+$	$+$
II	$90^\circ<\theta<180^\circ$	$+$	$-$	$-$
III	$180^\circ<\theta<270^\circ$	$-$	$-$	$+$
IV	$270^\circ<\theta<360^\circ$	$-$	$+$	$-$

This follows because sine is the $y$ coordinate, cosine is the $x$ coordinate, and tangent is $y/x$.

The reference angle is the acute angle between the terminal arm and the $x$ axis. Find this angle, use the first-quadrant table for the magnitude, then apply the correct sign.

Worked example 1: second quadrant

Find $\sin150^\circ$ and $\cos150^\circ$ exactly.

The reference angle is

$$180^\circ-150^\circ=30^\circ.$$

In quadrant II, sine is positive and cosine is negative. Hence

$$\boxed{\sin150^\circ=\sin30^\circ=\frac12},$$ $$\boxed{\cos150^\circ=-\cos30^\circ=-\frac{\sqrt3}{2}}.$$

Worked example 2: third quadrant in radians

Evaluate $\tan(5\pi/4)$ exactly.

Since $\pi$ radians is $180^\circ$,

$$\frac{5\pi}{4}=\pi+\frac{\pi}{4}.$$

The angle lies in quadrant III and its reference angle is $\pi/4$. Tangent is positive in quadrant III, so

$$\boxed{\tan\left(\frac{5\pi}{4}\right)=\tan\left(\frac{\pi}{4}\right)=1}.$$

Worked example 3: fourth quadrant

Find $\sin330^\circ$ exactly.

The reference angle is

$$360^\circ-330^\circ=30^\circ.$$

Sine is negative in quadrant IV. Therefore

$$\boxed{\sin330^\circ=-\sin30^\circ=-\frac12}.$$

Worked example 4: an angle outside one revolution

Evaluate $\cos(-315^\circ)$ exactly.

Add $360^\circ$ to find a coterminal angle:

$$-315^\circ+360^\circ=45^\circ.$$

Both angles finish at the same point on the unit circle, so

$$\boxed{\cos(-315^\circ)=\cos45^\circ=\frac{\sqrt2}{2}}.$$

Self check 2

Evaluate exactly:

1. $\cos120^\circ$
2. $\sin225^\circ$
3. $\tan(7\pi/6)$
4. $\cos(11\pi/6)$
5. $\sin(-60^\circ)$

Answers

1. Reference angle $60^\circ$, quadrant II: $-1/2$.
2. Reference angle $45^\circ$, quadrant III: $-\sqrt2/2$.
3. Reference angle $\pi/6$, quadrant III: $\sqrt3/3$.
4. Reference angle $\pi/6$, quadrant IV: $\sqrt3/2$.
5. $-60^\circ$ is coterminal with $300^\circ$, so the answer is $-\sqrt3/2$.

Combining exact values

Keep values as fractions and surds until the final line. Ordinary fraction arithmetic and careful surd simplification are then enough.

Worked example 5: an exact expression

Evaluate

$$2\sin60^\circ\cos30^\circ-\tan45^\circ.$$

Substitute exact values:

$$\begin{aligned} 2\sin60^\circ\cos30^\circ-\tan45^\circ &=2\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)-1\\ &=2\left(\frac34\right)-1\\ &=\frac32-1\\ &=\boxed{\frac12}. \end{aligned}$$

Worked example 6: solve using an exact value

Solve

$$4\cos x-\sqrt2=0, \qquad 0\leq x\leq2\pi.$$

First isolate the trigonometric function:

$$\cos x=\frac{\sqrt2}{4}=\frac1{2}\cdot\frac{\sqrt2}{2}.$$

This is not one of the standard exact cosine values, so $x$ is not a special angle. An inverse cosine would be required. By contrast, if the equation were $2\cos x-\sqrt2=0$, then

$$\cos x=\frac{\sqrt2}{2},$$

giving

$$x=\frac\pi4\quad\text{or}\quad x=\frac{7\pi}{4}.$$

Recognising when an exact table value does not apply is as important as recalling the table.

Common misconceptions

• $\sin^{-1}x$ means inverse sine, not $1/\sin x$. The reciprocal of sine is cosecant.
• $\sin30^\circ=1/2$, but $\sin(30)$ in radian mode means $\sin(30\text{ radians})$ and has a different value.
• $\sqrt3/2$ means $(\sqrt3)/2$, not $\sqrt{3/2}$.
• $\tan90^\circ$ is undefined because it requires division by zero.
• A reference angle gives the magnitude only. The quadrant determines the sign.
• Do not assume every multiple of $30^\circ$ or $45^\circ$ has a positive value.
• Avoid converting an exact surd to a decimal unless the question requests an approximation.

Check your understanding

1. Reconstruct the first-quadrant exact-value table without a calculator.
2. Explain geometrically why $\sin30^\circ=\cos60^\circ$.
3. Evaluate $3\cos(2\pi/3)-2\sin(7\pi/6)$ exactly.
4. Evaluate $\dfrac{\tan135^\circ}{\sin315^\circ}$ exactly.
5. Find all $x$ satisfying $\sin x=-\sqrt3/2$ for $0\leq x\leq2\pi$.
6. A student writes $\cos210^\circ=\sqrt3/2$. Identify and correct the error.

Answers

1. See the table near the start of this page. Sine follows $\sqrt0/2$ to $\sqrt4/2$, cosine reverses it, and tangent is sine divided by cosine.
2. In the same $30^\circ$, $60^\circ$, $90^\circ$ triangle, the side opposite $30^\circ$ is adjacent to $60^\circ$. Both ratios use the same hypotenuse.
3. $3(-1/2)-2(-1/2)=-3/2+1=\boxed{-1/2}$.
4. $\tan135^\circ=-1$ and $\sin315^\circ=-\sqrt2/2$, so the quotient is $2/\sqrt2=\boxed{\sqrt2}$.
5. The reference angle is $\pi/3$. Sine is negative in quadrants III and IV, so $\boxed{x=4\pi/3,\ 5\pi/3}$.
6. The reference angle is $30^\circ$, but $210^\circ$ is in quadrant III, where cosine is negative. Thus $\boxed{\cos210^\circ=-\sqrt3/2}$.

Next steps

Use these values when studying trigonometric graphs, trigonometric identities and trigonometric equations. They are also the starting values for deriving compound-angle and double-angle formulae.