Cos 5pi/6 - Exact Value -√3/2 and How to Find It

#Trigonometry
TL;DR
The value of cos 5pi/6 is exactly −32-\dfrac{\sqrt{3}}{2}, about −0.8660-0.8660. The angle 5π6\dfrac{5\pi}{6} lands in Quadrant II with a reference angle of π6\dfrac{\pi}{6}, where cosine is negative, so this article shows the reference-angle method, the unit circle proof, a standard-angle table, worked examples, and the common mistakes.
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Bhanzu TeamLast updated on August 11, 20265 min read

What Does Cos 5pi/6 Mean?

On the unit circle, the cosine of an angle is the xx-coordinate of the point where the angle's radius meets the circle. A quadrant is one of the four regions the axes cut the plane into, numbered anticlockwise from the top right.

The angle 5π6\dfrac{5\pi}{6} is measured anticlockwise from the positive xx-axis and stops in Quadrant II, the top-left region. Its terminal point is (−32,12)\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right), so the xx-coordinate, and therefore the cosine, is −32-\dfrac{\sqrt{3}}{2}. Points in Quadrant II have negative xx and positive yy, which is why cosine is negative here while sine is positive.

Where Does Cos 5pi/6 Show Up?

The angle 5π6\dfrac{5\pi}{6} points to 150∘150^\circ, which is 30∘30^\circ above the negative xx-axis, into the top-left. A strut leaning up and to the left, a vector aimed northwest of due-west, or a rotating arm just past the vertical all have a horizontal component read from cos⁡5π6=−32\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}. The value is negative because the point has crossed into the left half of the circle while staying above the horizontal.

Standard-Angle Reference Table

The value 32\dfrac{\sqrt{3}}{2} and its negative appear at four angles that share the reference angle π6\dfrac{\pi}{6}. The quadrant decides the sign.

Angle (radians)

Angle (degrees)

Quadrant

cos⁡θ\cos\theta

π6\dfrac{\pi}{6}

30∘30^\circ

I

32\dfrac{\sqrt{3}}{2}

5π6\dfrac{5\pi}{6}

150∘150^\circ

II

−32-\dfrac{\sqrt{3}}{2}

7π6\dfrac{7\pi}{6}

210∘210^\circ

III

−32-\dfrac{\sqrt{3}}{2}

11π6\dfrac{11\pi}{6}

330∘330^\circ

IV

32\dfrac{\sqrt{3}}{2}

All four have the same reference angle, so their cosines share the magnitude 32\dfrac{\sqrt{3}}{2}. Since 5π6\dfrac{5\pi}{6} sits in Quadrant II, its cosine is negative - the same magnitude as cos 30 degrees, with a minus sign added by the quadrant.

How Do You Find The Exact Value Of Cos 5pi/6?

Three routes all give −32-\dfrac{\sqrt{3}}{2}.

Method 1: Reference angle and quadrant sign.

A reference angle is the acute angle between the terminal side and the xx-axis. For a Quadrant II angle, subtract from π\pi:

π−5π6=6π−5π6=π6\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}

Cosine is negative in Quadrant II, so:

cos⁡5π6=−cos⁡π6=−32\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}

Method 2: The unit circle.

Convert with the radians-to-degrees rule: 5π6=150∘\dfrac{5\pi}{6} = 150^\circ. The terminal point at 150∘150^\circ is (−32,12)\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right), so

cos⁡5π6=x-coordinate=−32\cos\frac{5\pi}{6} = x\text{-coordinate} = -\frac{\sqrt{3}}{2}

Method 3: From the reference-angle twin.

The magnitude comes straight from cos π/6, which is 32\dfrac{\sqrt{3}}{2}. Quadrant II attaches the minus sign, giving −32-\dfrac{\sqrt{3}}{2}. This is why cos⁡5π6\cos\dfrac{5\pi}{6} and cos⁡π6\cos\dfrac{\pi}{6} have the same size but opposite signs.

Examples Of Cos 5pi/6

Example 1

Evaluate 2cos⁡5π62\cos\dfrac{5\pi}{6}.

2cos⁡5π6=2×(−32)=−3≈−1.7322\cos\frac{5\pi}{6} = 2 \times \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \approx -1.732

Example 2

Find cos⁡5π6\cos\dfrac{5\pi}{6} using the reference angle.

Wrong attempt. A student finds the reference angle π6\dfrac{\pi}{6}, reads cos⁡π6=32\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}, and writes cos⁡5π6=32\cos\dfrac{5\pi}{6} = \dfrac{\sqrt{3}}{2}.

That skips the quadrant step. The reference angle only sets the size of the value, and 5π6\dfrac{5\pi}{6} is in Quadrant II where the xx-coordinate is negative.

Correct. Apply the Quadrant II sign: cos⁡5π6=−32\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}.

Example 3

Compare cos⁡5π6\cos\dfrac{5\pi}{6} with sin⁡5π6\sin\dfrac{5\pi}{6}.

In Quadrant II, cosine is negative but sine is positive:

cos⁡5π6=−32,sin⁡5π6=12\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \qquad \sin\frac{5\pi}{6} = \frac{1}{2}

Example 4

Evaluate cos⁡5π6+cos⁡π6\cos\dfrac{5\pi}{6} + \cos\dfrac{\pi}{6}.

The two share a magnitude and have opposite signs:

−32+32=0-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 0

Example 5

Find sec⁡5π6\sec\dfrac{5\pi}{6}.

Secant is the reciprocal of cosine:

sec⁡5π6=1−32=−23=−233\sec\frac{5\pi}{6} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}

Where Students Trip Up On Cos 5pi/6

Mistake 1: Forgetting the Quadrant II minus sign

Where it slips in: Finding the reference angle, reading the positive value, and stopping there.

Don't do this: Writing cos⁡5π6=32\cos\dfrac{5\pi}{6} = \dfrac{\sqrt{3}}{2}. The reference-angle value is always positive, and handing in that number without the quadrant sign is the usual error.

The correct way: In Quadrant II the xx-coordinate is negative, so cosine is negative: cos⁡5π6=−32\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}. Pair every reference angle with a quadrant-sign check.

Mistake 2: Using the wrong reference angle

Where it slips in: Subtracting from 2π2\pi or from π2\dfrac{\pi}{2} instead of from π\pi.

Don't do this: Computing 5π6−π2=π3\dfrac{5\pi}{6} - \dfrac{\pi}{2} = \dfrac{\pi}{3} and treating that as the reference angle.

The correct way: For a Quadrant II angle, the reference angle is π\pi minus the angle: π−5π6=π6\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}.

Mistake 3: Swapping the cosine and sine signs in Quadrant II

Where it slips in: Remembering that "one of them is positive" in Quadrant II and guessing which.

Don't do this: Making sine negative and cosine positive, the reverse of the truth.

The correct way: In Quadrant II, sine (yy) is positive and cosine (xx) is negative. Direction bearings carry the same logic: 150∘150^\circ points up and to the left, so its horizontal reach is negative while its vertical reach is positive.

Key Takeaways

  • Cos 5pi/6 equals −32-\dfrac{\sqrt{3}}{2}, about −0.8660-0.8660 - negative because 5π6\dfrac{5\pi}{6} lies in Quadrant II.

  • The reference angle is π6\dfrac{\pi}{6}, so the magnitude matches cos⁡30∘=32\cos 30^\circ = \dfrac{\sqrt{3}}{2}; the quadrant supplies the minus sign.

  • In degrees, cos⁡5π6=cos⁡150∘=−32\cos\dfrac{5\pi}{6} = \cos 150^\circ = -\dfrac{\sqrt{3}}{2}, and the terminal point is (−32,12)\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right).

  • The most common slip is reporting the positive reference-angle value and forgetting the Quadrant II sign.

To master quadrant signs with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor, or structured math classes online.

Practice These Before Moving On

  1. Evaluate 4cos⁡5π6+sec⁡5π64\cos\dfrac{5\pi}{6} + \sec\dfrac{5\pi}{6}.

  2. Without a calculator, decide the sign of cos⁡7π6\cos\dfrac{7\pi}{6} and give its value.

  3. Show that cos⁡5π6+cos⁡π6=0\cos\dfrac{5\pi}{6} + \cos\dfrac{\pi}{6} = 0 using the reference angle.

Want a live Bhanzu trainer to walk through Quadrant II signs and reference angles? Book a free demo class.

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Frequently Asked Questions

Is cos 5pi/6 positive or negative?
Negative. The angle is in Quadrant II, where cosine (the xx-coordinate) is negative, so cos⁡5π6=−32\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}.
What is cos 5pi/6 in degrees?
5π6\dfrac{5\pi}{6} radians is 150∘150^\circ, and cos⁡150∘=−32\cos 150^\circ = -\dfrac{\sqrt{3}}{2}.
What is the reference angle for 5pi/6?
π6\dfrac{\pi}{6}, found by subtracting 5π6\dfrac{5\pi}{6} from π\pi. It is the same reference angle as 30∘30^\circ.
Why is cos 5pi/6 the negative of cos pi/6?
Because both share the reference angle π6\dfrac{\pi}{6}, but 5π6\dfrac{5\pi}{6} is in Quadrant II where cosine is negative, so its value is −32-\dfrac{\sqrt{3}}{2}.
✍️ Written By
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Bhanzu Team
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