Finding Angles from 0° to 360°

If $\sin\theta = \tfrac{1}{2}$, then $\theta$ could be $30^\circ$. But it could also be $150^\circ$. Between $0^\circ$ and $360^\circ$, most trig ratios come from two different angles. This page shows how to find both, which you’ll need for solving trig equations and for sinusoidal models. All angles are in degrees.

Key ideas

Why there are two angles

Two terminal arms in different quadrants can have the same reference angle. If the ratio has the same sign in both quadrants, both angles give the same value.

$30^\circ$ and $150^\circ$ both end at height $\tfrac{1}{2}$, so $\sin 30^\circ = \sin 150^\circ = \tfrac{1}{2}$.

Steps

1. Find the reference angle $\beta$ from the positive value of the ratio, using special angles or the calculator’s inverse key ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$).
2. Use the sign and CAST to decide which two quadrants the angles are in.
3. Place $\beta$ in each quadrant:

Quadrant	Angle
I	$\beta$
II	$180^\circ - \beta$
III	$180^\circ + \beta$
IV	$360^\circ - \beta$

Values on the axes

Ratios of $0$, $1$, or $-1$ (and undefined tangents) come from angles on the axes. For example, $\sin\theta = -1$ only at $\theta = 270^\circ$, and $\cos\theta = 0$ at $90^\circ$ and $270^\circ$.

Why the calculator isn’t enough

The inverse keys give only one angle. $\sin^{-1}$ and $\tan^{-1}$ answer between $-90^\circ$ and $90^\circ$; $\cos^{-1}$ answers between $0^\circ$ and $180^\circ$. Use the calculator only for the reference angle, with a positive input.

Worked examples

Example 1: A special value

Solve $\sin\theta = \tfrac{1}{2}$ for $0^\circ \le \theta \le 360^\circ$.

Solution. $\beta = 30^\circ$. Sine is positive in quadrants I and II:

$$\theta = 30^\circ \qquad \text{or} \qquad \theta = 180^\circ - 30^\circ = 150^\circ$$

Example 2: A negative cosine

Solve $\cos\theta = -\tfrac{\sqrt{3}}{2}$ for $0^\circ \le \theta \le 360^\circ$.

Solution. From $\cos\beta = \tfrac{\sqrt{3}}{2}$, $\beta = 30^\circ$. Cosine is negative in quadrants II and III:

$$\theta = 180^\circ - 30^\circ = 150^\circ \qquad \text{or} \qquad \theta = 180^\circ + 30^\circ = 210^\circ$$

Example 3: Using the calculator

Solve $\tan\theta = -1.2$ for $0^\circ \le \theta \le 360^\circ$, to the nearest tenth of a degree.

Solution. $\beta = \tan^{-1}(1.2) \approx 50.2^\circ$. Tangent is negative in quadrants II and IV:

$$\theta \approx 180^\circ - 50.2^\circ = 129.8^\circ \qquad \text{or} \qquad \theta \approx 360^\circ - 50.2^\circ = 309.8^\circ$$

Example 4: A negative sine

Solve $\sin\theta = -0.4$ for $0^\circ \le \theta \le 360^\circ$, to the nearest tenth of a degree.

Solution. $\beta = \sin^{-1}(0.4) \approx 23.6^\circ$. Sine is negative in quadrants III and IV:

$$\theta \approx 180^\circ + 23.6^\circ = 203.6^\circ \qquad \text{or} \qquad \theta \approx 360^\circ - 23.6^\circ = 336.4^\circ$$

Common mistakes

Stopping at one angle. Unless the value is $1$, $-1$, or $0$ for sine, cosine, or tangent, there are two answers between $0^\circ$ and $360^\circ$.

Putting a negative number into the inverse key. $\sin^{-1}(-0.4) \approx -23.6^\circ$ isn’t the reference angle. Use the positive value to get $\beta$, then let the sign choose the quadrants.

Using the wrong formula for the quadrant. Quadrant III is $180^\circ + \beta$ and quadrant IV is $360^\circ - \beta$. Sketch the terminal arm if you’re unsure.

Choosing quadrants from the wrong ratio. For $\cos\theta \lt 0$, look where cosine is negative (II and III), not sine.

Practice

1. (Warm-up) Solve $\cos\theta = \tfrac{1}{2}$ for $0^\circ \le \theta \le 360^\circ$.

Solution

$\beta = 60^\circ$, cosine positive in I and IV: $\theta = 60^\circ$ or $300^\circ$.

2. (Warm-up) Solve $\tan\theta = 1$ for $0^\circ \le \theta \le 360^\circ$.

Solution

$\beta = 45^\circ$, tangent positive in I and III: $\theta = 45^\circ$ or $225^\circ$.

3. (Warm-up) Solve $\sin\theta = -\tfrac{\sqrt{2}}{2}$ for $0^\circ \le \theta \le 360^\circ$.

Solution

$\beta = 45^\circ$, sine negative in III and IV: $\theta = 225^\circ$ or $315^\circ$.

4. (Core) Solve $\cos\theta = 0.25$ for $0^\circ \le \theta \le 360^\circ$, to the nearest tenth of a degree.

Solution

$\beta = \cos^{-1}(0.25) \approx 75.5^\circ$, cosine positive in I and IV: $\theta \approx 75.5^\circ$ or $284.5^\circ$.

5. (Core) Solve $\sin\theta = 0.8$ for $0^\circ \le \theta \le 360^\circ$, to the nearest tenth of a degree.

Solution

$\beta = \sin^{-1}(0.8) \approx 53.1^\circ$, sine positive in I and II: $\theta \approx 53.1^\circ$ or $126.9^\circ$.

6. (Core) Solve $\tan\theta = -\sqrt{3}$ for $0^\circ \le \theta \le 360^\circ$.

Solution

$\beta = 60^\circ$, tangent negative in II and IV: $\theta = 120^\circ$ or $300^\circ$.

7. (Core) Solve $\sin\theta = -1$ for $0^\circ \le \theta \le 360^\circ$.

Solution

Only the point $(0, -1)$ has $\tfrac{y}{r} = -1$, so $\theta = 270^\circ$ (just one answer).

8. (Core) Solve $2\cos\theta + 1 = 0$ for $0^\circ \le \theta \le 360^\circ$.

Solution

$\cos\theta = -\tfrac{1}{2}$, so $\beta = 60^\circ$, with cosine negative in II and III: $\theta = 120^\circ$ or $240^\circ$.

9. (Challenge) Solve $3\sin\theta - 2 = 0$ for $0^\circ \le \theta \le 360^\circ$, to the nearest tenth of a degree.

Solution

$\sin\theta = \tfrac{2}{3}$, so $\beta = \sin^{-1}\left(\tfrac{2}{3}\right) \approx 41.8^\circ$. Sine positive in I and II: $\theta \approx 41.8^\circ$ or $138.2^\circ$.

10. (Challenge) Solve $\sin\theta = \cos\theta$ for $0^\circ \le \theta \le 360^\circ$.

Solution

Where $\cos\theta \ne 0$, divide both sides by $\cos\theta$: $\tan\theta = 1$, so $\theta = 45^\circ$ or $225^\circ$. (At $90^\circ$ and $270^\circ$, $\cos\theta = 0$ but $\sin\theta = \pm 1$, so they aren’t solutions.)

Check: $\sin 225^\circ = \cos 225^\circ = -\tfrac{\sqrt{2}}{2}$. ✓