Find the exact values of $\sin \frac{\pi}{6}$, $\cos \frac{\pi}{4}$ and $\tan \frac{\pi}{3}$.

$\sin \frac{\pi}{6} = \frac{1}{2}$, $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\tan \frac{\pi}{3} = \sqrt{3}$

Find the exact value of $\cos \frac{5\pi}{6}$.

$\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$

Find the exact values of $\tan \frac{7\pi}{4}$ and $\csc \frac{4\pi}{3}$.

$\tan \frac{7\pi}{4} = -1$ and $\csc \frac{4\pi}{3} = -\frac{2\sqrt{3}}{3}$

The point $P\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ is on the unit circle at angle $\theta$, with $0 \le \theta < 2\pi$. Find $\theta$ and $\tan \theta$.

$\theta = \frac{7\pi}{6}$ and $\tan \theta = \frac{\sqrt{3}}{3}$

Find every $\theta$ in $0 \le \theta < 2\pi$ with $\cos \theta = -\frac{1}{2}$, the point on the unit circle at each, and the exact value of $\tan \theta$ at each.

$\theta = \frac{2\pi}{3}$ at $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ with $\tan \theta = -\sqrt{3}$, and $\theta = \frac{4\pi}{3}$ at $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ with $\tan \theta = \sqrt{3}$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Unit Circle and Special Angles

On a circle of radius $1$ , the point at angle $\theta$ has coordinates $(\cos \theta, \sin \theta)$ . That single picture is where every exact trig value comes from, and it is why you can find $\sin \frac{7\pi}{6}$ without a calculator.

How It Works

Two triangles supply every value. Half a square, with sides $1, 1, \sqrt{2}$ , gives the $\frac{\pi}{4}$ ratios: $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan \frac{\pi}{4} = 1$ . Half an equilateral triangle, with sides $1, \sqrt{3}, 2$ , gives $\frac{\pi}{6}$ and $\frac{\pi}{3}$ : $\sin \frac{\pi}{6} = \frac{1}{2}$ , $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ , and the sine and cosine swap places for $\frac{\pi}{3}$ .

The axes supply the rest. At $0$ , $\frac{\pi}{2}$ , $\pi$ and $\frac{3\pi}{2}$ the point is $(1, 0)$ , $(0, 1)$ , $(-1, 0)$ and $(0, -1)$ , so the cosine and sine are read straight off the coordinates, and tangent is undefined wherever the cosine is $0$ .

Any other special angle has a reference angle. It is the acute angle between the terminal arm and the $x$ -axis: $\frac{5\pi}{6}$ is $\frac{\pi}{6}$ short of $\pi$ , so its reference angle is $\frac{\pi}{6}$ . The reference angle fixes the size of each ratio; the quadrant fixes the sign.

CAST gives the sign. Starting in quadrant 4 and going counterclockwise: Cosine positive in quadrant 4, All positive in quadrant 1, Sine positive in quadrant 2, Tangent positive in quadrant 3. Equivalently, the sign of the cosine is the sign of $x$ , the sign of the sine is the sign of $y$ , and tangent is $\frac{y}{x}$ .

The reciprocal ratios follow: $\csc \theta = \frac{1}{\sin \theta}$ , $\sec \theta = \frac{1}{\cos \theta}$ , $\cot \theta = \frac{1}{\tan \theta}$ . Flip the exact value, then rationalise the denominator, so $\frac{2}{\sqrt{3}}$ is written $\frac{2\sqrt{3}}{3}$ .

Working an angle out means three questions in order: which quadrant, which reference angle, which sign. Skip the first and the sign goes wrong on half of every test.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the exact values of $\sin \frac{\pi}{6}$ , $\cos \frac{\pi}{4}$ and $\tan \frac{\pi}{3}$ .
Show solution
1. $\frac{\pi}{6}$ is $30^\circ$ . In the $1, \sqrt{3}, 2$ triangle the side opposite $30^\circ$ is $1$ , so $\sin \frac{\pi}{6} = \frac{1}{2}$ .
2. $\frac{\pi}{4}$ is $45^\circ$ . In the $1, 1, \sqrt{2}$ triangle, $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .
3. $\frac{\pi}{3}$ is $60^\circ$ : opposite $\sqrt{3}$ , adjacent $1$ , so $\tan \frac{\pi}{3} = \sqrt{3}$ .
Answer: $\sin \frac{\pi}{6} = \frac{1}{2}$ , $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , $\tan \frac{\pi}{3} = \sqrt{3}$
Easy
Find the exact value of $\cos \frac{5\pi}{6}$ .
Show solution
1. $\frac{5\pi}{6}$ is between $\frac{\pi}{2}$ and $\pi$ , so the terminal arm is in quadrant 2.
2. Reference angle: $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$ , and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ .
3. In quadrant 2 only sine is positive, so the cosine is negative.
Answer: $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$
Medium
Find the exact values of $\tan \frac{7\pi}{4}$ and $\csc \frac{4\pi}{3}$ .
Show solution
1. $\frac{7\pi}{4}$ is $\frac{\pi}{4}$ short of $2\pi$ : quadrant 4, reference angle $\frac{\pi}{4}$ . Tangent is negative there, so $\tan \frac{7\pi}{4} = -1$ .
2. $\frac{4\pi}{3}$ is $\frac{\pi}{3}$ past $\pi$ : quadrant 3, reference angle $\frac{\pi}{3}$ . Sine is negative there: $\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$ .
3. Flip it: $\csc \frac{4\pi}{3} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$ .
Answer: $\tan \frac{7\pi}{4} = -1$ and $\csc \frac{4\pi}{3} = -\frac{2\sqrt{3}}{3}$
Medium
The point $P\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ is on the unit circle at angle $\theta$ , with $0 \le \theta < 2\pi$ . Find $\theta$ and $\tan \theta$ .
Show solution
1. Both coordinates are negative, so $P$ is in quadrant 3.
2. $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \theta = -\frac{1}{2}$ have the sizes of the $\frac{\pi}{6}$ ratios, so the reference angle is $\frac{\pi}{6}$ .
3. Quadrant 3: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ .
4. $\tan \theta = \frac{y}{x} = \frac{-1/2}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ , positive, as tangent should be in quadrant 3.
Answer: $\theta = \frac{7\pi}{6}$ and $\tan \theta = \frac{\sqrt{3}}{3}$
Test level
Find every $\theta$ in $0 \le \theta < 2\pi$ with $\cos \theta = -\frac{1}{2}$ , the point on the unit circle at each, and the exact value of $\tan \theta$ at each.
Show solution
1. $\cos \theta$ is negative in quadrants 2 and 3, and $\frac{1}{2}$ is the cosine of $\frac{\pi}{3}$ , so the reference angle is $\frac{\pi}{3}$ .
2. Quadrant 2: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . Quadrant 3: $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$ .
3. The points are $(\cos \theta, \sin \theta)$ : $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ at $\frac{2\pi}{3}$ and $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ at $\frac{4\pi}{3}$ .
4. $\tan \theta = \frac{y}{x}$ : at $\frac{2\pi}{3}$ it is $\frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$ ; at $\frac{4\pi}{3}$ it is $\frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$ .
Answer: $\theta = \frac{2\pi}{3}$ at $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ with $\tan \theta = -\sqrt{3}$ , and $\theta = \frac{4\pi}{3}$ at $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ with $\tan \theta = \sqrt{3}$

Learn It on Martian Lab

Martian Lab teaches this in the Trigonometry lesson of its MHF4U Trigonometric Ratios unit. Watch it, then take the Trigonometric Ratios practice tests to see if it stuck.

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More Trigonometric Ratios topics: Proving Trig Identities · Radian Measure · Compound Angle Formulas · Double Angle Formulas

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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