Trig Ratios of Any Angle

So far, sine, cosine, and tangent have only applied to acute angles in right triangles. By placing angles on a coordinate grid, we can give them meaning for any angle from $0^\circ$ to $360^\circ$. This is what makes the sinusoidal functions later in the course possible. All angles are in degrees.

Key ideas

Angles in standard position

An angle is in standard position when its vertex is at the origin and its initial arm lies along the positive $x$-axis. The terminal arm rotates counterclockwise from there.

Ratios from a point on the terminal arm

If $P(x, y)$ is any point on the terminal arm, and $r$ is its distance from the origin, $r = \sqrt{x^2 + y^2}$, then:

$$\sin\theta = \frac{y}{r} \qquad \cos\theta = \frac{x}{r} \qquad \tan\theta = \frac{y}{x}$$

Since $r$ is always positive, the signs of the ratios depend on the signs of $x$ and $y$, which means on the quadrant.

The CAST rule

Quadrant	Angles	Positive ratios
I	$0^\circ$ to $90^\circ$	All
II	$90^\circ$ to $180^\circ$	Sine
III	$180^\circ$ to $270^\circ$	Tangent
IV	$270^\circ$ to $360^\circ$	Cosine

Reading the quadrants in order IV, I, II, III spells CAST. The ratios not listed are negative in that quadrant.

Reference angles

The reference angle $\beta$ is the acute angle between the terminal arm and the $x$-axis:

Quadrant	Reference angle
I	$\beta = \theta$
II	$\beta = 180^\circ - \theta$
III	$\beta = \theta - 180^\circ$
IV	$\beta = 360^\circ - \theta$

Any trig ratio of $\theta$ equals the same ratio of $\beta$, with the sign from CAST. For example, $\sin 150^\circ = +\sin 30^\circ = \tfrac{1}{2}$.

Angles on the axes

For $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$, use a point with $r = 1$, such as $(0, -1)$ for $270^\circ$. Then $\sin 270^\circ = -1$, $\cos 270^\circ = 0$, and $\tan 270^\circ$ is undefined.

Worked examples

Example 1: From a point

The point $P(-3, 4)$ is on the terminal arm of $\theta$. Find the three primary ratios, and $\theta$ to the nearest tenth of a degree.

Solution. $r = \sqrt{(-3)^2 + 4^2} = 5$.

$$\sin\theta = \frac{4}{5}, \qquad \cos\theta = -\frac{3}{5}, \qquad \tan\theta = -\frac{4}{3}$$

The reference angle is $\beta = \tan^{-1}\left(\tfrac{4}{3}\right) \approx 53.1^\circ$. $P$ is in quadrant II, so $\theta = 180^\circ - 53.1^\circ \approx 126.9^\circ$.

Example 2: Exact values with reference angles

Find the exact values of $\sin 150^\circ$, $\cos 225^\circ$, and $\tan 300^\circ$.

Solution.

• $150^\circ$ is in quadrant II, $\beta = 30^\circ$, and sine is positive there: $\sin 150^\circ = \tfrac{1}{2}$.
• $225^\circ$ is in quadrant III, $\beta = 45^\circ$, and cosine is negative there: $\cos 225^\circ = -\tfrac{\sqrt{2}}{2}$.
• $300^\circ$ is in quadrant IV, $\beta = 60^\circ$, and tangent is negative there: $\tan 300^\circ = -\sqrt{3}$.

Example 3: Angles on the axes

Find $\sin 270^\circ$, $\cos 180^\circ$, and $\tan 90^\circ$.

Solution. Use points with $r = 1$: $(0, -1)$ for $270^\circ$, $(-1, 0)$ for $180^\circ$, and $(0, 1)$ for $90^\circ$.

$$\sin 270^\circ = \frac{-1}{1} = -1, \qquad \cos 180^\circ = \frac{-1}{1} = -1, \qquad \tan 90^\circ = \frac{1}{0} \text{ (undefined)}$$

Example 4: From one ratio to the others

If $\cos\theta = -\tfrac{5}{13}$ and $\theta$ is in quadrant III, find $\sin\theta$ and $\tan\theta$.

Solution. Use $x = -5$ and $r = 13$. In quadrant III, $y$ is negative:

$$y = -\sqrt{13^2 - 5^2} = -\sqrt{144} = -12$$ $$\sin\theta = -\frac{12}{13}, \qquad \tan\theta = \frac{-12}{-5} = \frac{12}{5}$$

Check with CAST: in quadrant III, only tangent is positive. ✓

Common mistakes

Measuring the reference angle from the $y$-axis. The reference angle is always measured to the $x$-axis.

Forgetting the sign. The reference angle gives the size of the ratio; CAST gives the sign. $\cos 225^\circ$ is $-\tfrac{\sqrt{2}}{2}$, not $\tfrac{\sqrt{2}}{2}$.

Giving $r$ a sign. $r$ is a distance, so it’s always positive. Only $x$ and $y$ can be negative.

Trusting the calculator’s inverse for any quadrant. $\tan^{-1}\left(-\tfrac{4}{3}\right)$ gives about $-53.1^\circ$, which isn’t the angle in Example 1. Find the reference angle, then place it in the correct quadrant.

Practice

1. (Warm-up) Which quadrant is each angle in, and which primary ratio is positive there?

• (a) $200^\circ$
• (b) $310^\circ$
• (c) $95^\circ$

Solution

(a) Quadrant III: tangent.

(b) Quadrant IV: cosine.

(c) Quadrant II: sine.

2. (Warm-up) Find the reference angle for $160^\circ$, $250^\circ$, and $335^\circ$.

Solution

$180^\circ - 160^\circ = 20^\circ$; $250^\circ - 180^\circ = 70^\circ$; $360^\circ - 335^\circ = 25^\circ$.

3. (Warm-up) The point $(5, -12)$ is on the terminal arm of $\theta$. Find $\sin\theta$, $\cos\theta$, and $\tan\theta$.

Solution

$r = \sqrt{25 + 144} = 13$.

$\sin\theta = -\tfrac{12}{13}$, $\cos\theta = \tfrac{5}{13}$, $\tan\theta = -\tfrac{12}{5}$.

4. (Core) Find the exact values of $\cos 120^\circ$, $\sin 315^\circ$, and $\tan 210^\circ$.

Solution

$\cos 120^\circ = -\tfrac{1}{2}$ (quadrant II, $\beta = 60^\circ$).

$\sin 315^\circ = -\tfrac{\sqrt{2}}{2}$ (quadrant IV, $\beta = 45^\circ$).

$\tan 210^\circ = \tfrac{\sqrt{3}}{3}$ (quadrant III, $\beta = 30^\circ$, tangent positive).

5. (Core) Find $\sin 180^\circ$, $\cos 270^\circ$, and $\tan 360^\circ$.

Solution

All three are $0$. ($180^\circ$ uses the point $(-1, 0)$, $270^\circ$ uses $(0, -1)$, and $360^\circ$ uses $(1, 0)$.)

6. (Core) If $\tan\theta = -\tfrac{3}{4}$ and $\theta$ is in quadrant II, find $\sin\theta$ and $\cos\theta$.

Solution

In quadrant II, $x \lt 0$ and $y \gt 0$, so use $x = -4$, $y = 3$, and $r = 5$.

$\sin\theta = \tfrac{3}{5}$, $\cos\theta = -\tfrac{4}{5}$.

7. (Core) The point $(-2, -2)$ is on the terminal arm of $\theta$. Find the exact primary ratios and the angle $\theta$.

Solution

$r = \sqrt{4 + 4} = 2\sqrt{2}$.

$\sin\theta = \tfrac{-2}{2\sqrt{2}} = -\tfrac{\sqrt{2}}{2}$, $\cos\theta = -\tfrac{\sqrt{2}}{2}$, $\tan\theta = 1$.

The point is in quadrant III with $\beta = 45^\circ$, so $\theta = 225^\circ$.

8. (Challenge) Explain why $\sin\theta$ can never be greater than $1$ or less than $-1$.

Solution

$\sin\theta = \tfrac{y}{r}$, and the distance $r = \sqrt{x^2 + y^2}$ is always at least as big as $|y|$. So $\tfrac{|y|}{r} \le 1$, which means $-1 \le \sin\theta \le 1$. The same argument works for cosine.

9. (Challenge) Show that $\sin^2 150^\circ + \cos^2 150^\circ = 1$ using exact values.

Solution

$\sin 150^\circ = \tfrac{1}{2}$ and $\cos 150^\circ = -\tfrac{\sqrt{3}}{2}$:

$$\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$