Cos 330 Degrees: Exact Value, Unit Circle And Examples

Where Does Cos 330 Degrees Show Up, And Why Does It Matter?

Any time a quantity points $30^\circ$ below the positive horizontal, its horizontal share is set by $\cos 330^\circ$. A rope pulling a crate with 20 newtons at $330^\circ$ delivers $20 \times \frac{\sqrt{3}}{2} \approx 17.32$ newtons along the floor, the standard way of splitting vectors into parts.

Computer graphics and robotics describe a clockwise turn of $30^\circ$ as a turn of $330^\circ$ (or $-30^\circ$), and both use $\cos 330^\circ = \frac{\sqrt{3}}{2}$ inside the rotation step. The value also marks one of the twelve equally spaced points of a regular dodecagon drawn on a unit circle, which is why it sits in every standard-angle table.

Standard-Angle Cosine Reference Table

Cosine at every multiple of $30^\circ$ around the full circle is listed below, so you can see $330^\circ$ next to its neighbours. Notice how $330^\circ$ repeats the value at $30^\circ$.

Only four magnitudes ever appear in this column: $1$, $\frac{\sqrt{3}}{2}$, $\frac{1}{2}$ and $0$. The quadrant decides the sign, and that single idea carries the whole article.

What Do Quadrant, Reference Angle And Coterminal Angle Mean For 330 Degrees?

A quadrant is one of the four regions the x- and y-axes cut the plane into, numbered I to IV counterclockwise from the top right. An angle of $330^\circ$, measured counterclockwise from the positive x-axis, lands in Quadrant IV, between $270^\circ$ and $360^\circ$.

The reference angle is the acute angle between the terminal side and the x-axis. For a Quadrant IV angle it is $360^\circ - \theta$, so the reference angle of $330^\circ$ is $30^\circ$.

Coterminal angles end on the same terminal side, so they differ by whole turns of $360^\circ$. That makes $330^\circ$, $-30^\circ$ and $690^\circ$ the same position, and all three share one cosine.

Cosine itself has two equal definitions. In a right triangle it is adjacent over hypotenuse; on the unit circle, a circle of radius 1 centred at the origin, it is the x-coordinate of the point the angle reaches.

How Do You Find The Exact Value Of Cos 330 Degrees?

Find the reference angle, take the cosine of that acute angle, then attach the sign of the quadrant. Every other route below reaches the same $\frac{\sqrt{3}}{2}$, and seeing them side by side is the best proof that the answer is right.

Method 1: Reference Angle And The ASTC Sign Rule

The ASTC rule ("All Students Take Calculus") lists which ratios are positive in Quadrants I, II, III and IV: All, Sine, Tangent, Cosine. Quadrant IV is the "C" quadrant, so cosine is positive there.

1. Locate the angle: $270^\circ < 330^\circ < 360^\circ$, so Quadrant IV.

2. Reference angle: $360^\circ - 330^\circ = 30^\circ$.

3. Value of the acute angle: cos 30 degrees is $\frac{\sqrt{3}}{2}$.

4. Sign: cosine is positive in Quadrant IV.

$$\cos 330^\circ = +\cos 30^\circ = \frac{\sqrt{3}}{2}$$

Method 2: Reading The Unit Circle

Turning $330^\circ$ counterclockwise lands the radius $30^\circ$ below the positive x-axis. The point there is $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$: the same horizontal distance as the $30^\circ$ point, mirrored below the axis.

$$\cos 330^\circ = x\text{-coordinate} = \frac{\sqrt{3}}{2}$$

That answers the common search "what is cos 330 on the unit circle": it is the x-coordinate $\frac{\sqrt{3}}{2}$, while the y-coordinate $-\frac{1}{2}$ is $\sin 330^\circ$.

Method 3: The Identity $\cos(360^\circ - \theta) = \cos\theta$

Subtracting an angle from a full turn reflects the point across the x-axis, which leaves the x-coordinate unchanged.

$$\cos 330^\circ = \cos(360^\circ - 30^\circ)$$

$$= \cos 30^\circ = \frac{\sqrt{3}}{2}$$

Method 4: Cosine Is An Even Function

Because $330^\circ$ and $-30^\circ$ are coterminal, $\cos 330^\circ = \cos(-30^\circ)$. Cosine is an even function, meaning $\cos(-\theta) = \cos\theta$, so the value is $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

Method 5: The Angle Sum Formula As A Check

You could start here, with $330^\circ = 270^\circ + 60^\circ$ and the sum and difference formulas. Hold on, though: for $330^\circ$ the reference angle already gave the answer in one line, so treat this route as an independent check rather than the main method.

$$\cos(270^\circ + 60^\circ) = \cos 270^\circ \cos 60^\circ - \sin 270^\circ \sin 60^\circ$$

$$= 0 \times \frac{1}{2} - (-1) \times \frac{\sqrt{3}}{2}$$

$$= \frac{\sqrt{3}}{2}$$

Method 6: Calculator Check In Degree Mode

Set the calculator to DEG, enter $\cos(330)$, and it returns $0.8660254038$. Squaring that gives $0.75 = \frac{3}{4}$, and $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$, so the decimal confirms the surd.

What Is Cos 330 Degrees In Radians?

In radians, $330^\circ$ is $\frac{11\pi}{6}$, so $\cos \frac{11\pi}{6} = \frac{\sqrt{3}}{2}$. The conversion multiplies degrees by $\frac{\pi}{180}$, the rule behind converting degrees to radians.

$$330^\circ \times \frac{\pi}{180^\circ} = \frac{330\pi}{180} = \frac{11\pi}{6} \approx 5.7596$$

A radian is the angle that cuts off an arc equal in length to the radius, so $\frac{11\pi}{6}$ radians means the arc is about $5.76$ radii long. In radian language the reference angle is $2\pi - \frac{11\pi}{6} = \frac{\pi}{6}$, which matches the $30^\circ$ found above.

What Are The Other Trig Ratios Of 330 Degrees?

All six ratios of $330^\circ$ come from the unit-circle point $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$. Socratic's popular question "How do you find the 6 trig functions of 330 degrees?" is answered by the table below, where only cosine and secant are positive.

Written in terms of its partners, $\cos 330^\circ = \frac{1}{\sec 330^\circ}$ by the reciprocal identities. It also equals $\sqrt{1 - \sin^2 330^\circ} = \sqrt{1 - \frac{1}{4}}$, taking the positive root because cosine is positive in Quadrant IV.

What Does The Cosine Graph Show At 330 Degrees?

On the graph of $y = \cos x$, the curve at $330^\circ$ is climbing back toward its maximum of 1 at $360^\circ$, and its height there is $0.8660$. The cosine function has period $360^\circ$, so the same height returns at $690^\circ$ and at $-30^\circ$.

The curve is symmetric about $x = 180^\circ$, which is the graph's version of $\cos(360^\circ - \theta) = \cos\theta$. Points the same distance either side of $180^\circ$ always share a height.

Examples Of Cos 330 Degrees

Example 1

Evaluate $4\cos 330^\circ$.

$$4\cos 330^\circ = 4 \times \frac{\sqrt{3}}{2}$$

$$= 2\sqrt{3}$$

Final answer: $2\sqrt{3} \approx 3.4641$

Example 2

A student writes $\cos 330^\circ = \cos(360^\circ - 30^\circ) = -\cos 30^\circ$. Find the correct value.

The tempting step treats the minus sign inside the angle as a minus sign on the answer, which gives $-\frac{\sqrt{3}}{2} \approx -0.8660$.

That cannot be right. The point at $330^\circ$ lies to the right of the y-axis, so its x-coordinate, the cosine, must be positive.

The rescue is the reflection identity: $\cos(360^\circ - \theta) = \cos\theta$, with no sign change.

$$\cos 330^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$$

Final answer: $\frac{\sqrt{3}}{2} \approx 0.8660$

Example 3

Find $\cos 330^\circ \cos 30^\circ + \sin 330^\circ \sin 30^\circ$.

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

$$= \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$$

The expression is $\cos(330^\circ - 30^\circ) = \cos 300^\circ$, and $\cos 300^\circ = \frac{1}{2}$ agrees.

Final answer: $\frac{1}{2}$

Example 4

A 20 N force acts at $330^\circ$. Find its horizontal and vertical components.

$$F_x = 20\cos 330^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$$

$$F_y = 20\sin 330^\circ = 20 \times \left(-\frac{1}{2}\right) = -10$$

Final answer: $F_x = 10\sqrt{3} \approx 17.32$ N to the right, $F_y = -10$ N (downward)

Example 5

Find $\cos(-390^\circ)$ using $\cos 330^\circ$.

Add two full turns: $-390^\circ + 720^\circ = 330^\circ$, so the angles are coterminal.

$$\cos(-390^\circ) = \cos 330^\circ = \frac{\sqrt{3}}{2}$$

Final answer: $\frac{\sqrt{3}}{2}$

Example 6

Use $\cos 330^\circ$ and the half-angle formula to find $\cos 165^\circ$.

Since $165^\circ = \frac{330^\circ}{2}$ lies in Quadrant II, where cosine is negative, take the negative root of the half angle formula.

$$\cos 165^\circ = -\sqrt{\frac{1 + \cos 330^\circ}{2}}$$

$$= -\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = -\frac{\sqrt{2 + \sqrt{3}}}{2}$$

$$= -\frac{\sqrt{6} + \sqrt{2}}{4}$$

Final answer: $-\frac{\sqrt{6} + \sqrt{2}}{4} \approx -0.9659$

The sign choice in this last step is where most errors creep in: the half-angle formula hands you $\pm$, and only the quadrant of $165^\circ$ can choose.

Where Do Students Go Wrong With Cos 330 Degrees?

Mistake 1: Calculator In Radian Mode

Where it slips in: Checking the answer on a calculator or phone that is set to radians.

Don't do this: Accept $\cos(330) = -0.9912$, which is the cosine of 330 radians, not 330 degrees. Google's own suggestion box shows this same $-0.991198822$ for some phrasings of the query.

The correct way: Switch to DEG, or enter $\frac{11\pi}{6}$ in radian mode; both return $0.8660$. A negative answer for a Quadrant IV cosine is the first sign that the mode is wrong.

The rusher is most exposed here: a number on a screen feels like proof. Students who say the quadrant out loud before pressing any key catch the $-0.9912$ immediately, because they already expect a positive value near $0.87$.

Mistake 2: Mixing Up The Sine And Cosine Of 330 Degrees

Where it slips in: Writing all six ratios quickly, when $-\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ get swapped.

Don't do this: Write $\cos 330^\circ = -\frac{1}{2}$. That is $\sin 330^\circ$.

The correct way: Cosine is the x-coordinate and sine is the y-coordinate of $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$. The memorizer who learned "330 means negative one half" without the picture makes this swap most often; sketching the point takes five seconds and settles it.

Mistake 3: Assuming An Inverse Cosine Gives Back 330 Degrees

Where it slips in: Solving $\cos\theta = \frac{\sqrt{3}}{2}$ and expecting the calculator to return $330^\circ$.

Don't do this: Claim $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 330^\circ$.

The correct way: The inverse cosine returns only its principal value, the single answer in the range $0^\circ$ to $180^\circ$, so it gives $30^\circ$. On $0^\circ$ to $360^\circ$ the full solution set is $\theta = 30^\circ$ or $\theta = 330^\circ$, and you add the second answer yourself.

Precision matters in the same way outside the classroom.

Where Can You Get Help With Trigonometry Values?

If reference angles and quadrant signs still feel shaky, working through them live with a trigonometry tutor is the fastest fix. Students preparing for later units can also book a high school math tutor or join regular math classes online.

Key Takeaways

• Cos 330 degrees equals $\frac{\sqrt{3}}{2} \approx 0.8660$, the same value as $\cos 30^\circ$.

• The angle lies in Quadrant IV with a $30^\circ$ reference angle, and cosine is positive there.

• In radians, $\cos \frac{11\pi}{6} = \frac{\sqrt{3}}{2}$, and the reference angle is $\frac{\pi}{6}$.

• The unit-circle point is $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$, so $\sin 330^\circ = -\frac{1}{2}$ and $\tan 330^\circ = -\frac{\sqrt{3}}{3}$.

• A calculator showing $-0.9912$ is in radian mode.

Practice These Before Moving On

1. Evaluate $2\cos 330^\circ + \sin 330^\circ$ in exact form.

2. A boat moves at 12 km/h on a heading of $330^\circ$ measured counterclockwise from east. Find its eastward speed.

3. Show that $\cos^2 330^\circ - \sin^2 330^\circ = \frac{1}{2}$, then name the angle whose cosine this is.

4. Find $\cos 690^\circ$ without a calculator.

Answers To The Practice Questions

1. $2 \times \frac{\sqrt{3}}{2} - \frac{1}{2} = \sqrt{3} - \frac{1}{2} \approx 1.2321$

2. $12\cos 330^\circ = 6\sqrt{3} \approx 10.39$ km/h

3. $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$, which is $\cos 660^\circ = \cos 300^\circ$ by the double angle formula

4. $690^\circ - 360^\circ = 330^\circ$, so $\cos 690^\circ = \frac{\sqrt{3}}{2}$

If question 3 stalls, return to Method 3 and the reflection idea. Want a trainer to walk through more problems like these with you? Book a free demo class.

Read More

• Trigonometric table of standard angles

• Trigonometric ratios of specific angles

• Cos 5π/3 and the 300 degree neighbour

• Cos 7π/4 in Quadrant IV

• Pythagorean identities