What Does Cos 5pi/6 Mean?
On the unit circle, the cosine of an angle is the $x$-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 $\dfrac{5\pi}{6}$ is measured anticlockwise from the positive $x$-axis and stops in Quadrant II, the top-left region. Its terminal point is $\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)$, so the $x$-coordinate, and therefore the cosine, is $-\dfrac{\sqrt{3}}{2}$. Points in Quadrant II have negative $x$ and positive $y$, which is why cosine is negative here while sine is positive.
Where Does Cos 5pi/6 Show Up?
The angle $\dfrac{5\pi}{6}$ points to $150^\circ$, which is $30^\circ$ above the negative $x$-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 $\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$. 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 $\dfrac{\sqrt{3}}{2}$ and its negative appear at four angles that share the reference angle $\dfrac{\pi}{6}$. The quadrant decides the sign.
Angle (radians) | Angle (degrees) | Quadrant | $\cos\theta$ |
|---|---|---|---|
$\dfrac{\pi}{6}$ | $30^\circ$ | I | $\dfrac{\sqrt{3}}{2}$ |
$\dfrac{5\pi}{6}$ | $150^\circ$ | II | $-\dfrac{\sqrt{3}}{2}$ |
$\dfrac{7\pi}{6}$ | $210^\circ$ | III | $-\dfrac{\sqrt{3}}{2}$ |
$\dfrac{11\pi}{6}$ | $330^\circ$ | IV | $\dfrac{\sqrt{3}}{2}$ |
All four have the same reference angle, so their cosines share the magnitude $\dfrac{\sqrt{3}}{2}$. Since $\dfrac{5\pi}{6}$ 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 $-\dfrac{\sqrt{3}}{2}$.
Method 1: Reference angle and quadrant sign.
A reference angle is the acute angle between the terminal side and the $x$-axis. For a Quadrant II angle, subtract from $\pi$:
$$\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}$$
Cosine is negative in Quadrant II, so:
$$\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$$
Method 2: The unit circle.
Convert with the radians-to-degrees rule: $\dfrac{5\pi}{6} = 150^\circ$. The terminal point at $150^\circ$ is $\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)$, so
$$\cos\frac{5\pi}{6} = x\text{-coordinate} = -\frac{\sqrt{3}}{2}$$
Method 3: From the reference-angle twin.
The magnitude comes straight from cos π/6, which is $\dfrac{\sqrt{3}}{2}$. Quadrant II attaches the minus sign, giving $-\dfrac{\sqrt{3}}{2}$. This is why $\cos\dfrac{5\pi}{6}$ and $\cos\dfrac{\pi}{6}$ have the same size but opposite signs.
Examples Of Cos 5pi/6
Example 1
Evaluate $2\cos\dfrac{5\pi}{6}$.
$$2\cos\frac{5\pi}{6} = 2 \times \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \approx -1.732$$
Example 2
Find $\cos\dfrac{5\pi}{6}$ using the reference angle.
Wrong attempt. A student finds the reference angle $\dfrac{\pi}{6}$, reads $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$, and writes $\cos\dfrac{5\pi}{6} = \dfrac{\sqrt{3}}{2}$.
That skips the quadrant step. The reference angle only sets the size of the value, and $\dfrac{5\pi}{6}$ is in Quadrant II where the $x$-coordinate is negative.
Correct. Apply the Quadrant II sign: $\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$.
Example 3
Compare $\cos\dfrac{5\pi}{6}$ with $\sin\dfrac{5\pi}{6}$.
In Quadrant II, cosine is negative but sine is positive:
$$\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \qquad \sin\frac{5\pi}{6} = \frac{1}{2}$$
Example 4
Evaluate $\cos\dfrac{5\pi}{6} + \cos\dfrac{\pi}{6}$.
The two share a magnitude and have opposite signs:
$$-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 0$$
Example 5
Find $\sec\dfrac{5\pi}{6}$.
Secant is the reciprocal of cosine:
$$\sec\frac{5\pi}{6} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$
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 $\cos\dfrac{5\pi}{6} = \dfrac{\sqrt{3}}{2}$. 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 $x$-coordinate is negative, so cosine is negative: $\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}$. Pair every reference angle with a quadrant-sign check.
Mistake 2: Using the wrong reference angle
Where it slips in: Subtracting from $2\pi$ or from $\dfrac{\pi}{2}$ instead of from $\pi$.
Don't do this: Computing $\dfrac{5\pi}{6} - \dfrac{\pi}{2} = \dfrac{\pi}{3}$ and treating that as the reference angle.
The correct way: For a Quadrant II angle, the reference angle is $\pi$ minus the angle: $\pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$.
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 ($y$) is positive and cosine ($x$) is negative. Direction bearings carry the same logic: $150^\circ$ points up and to the left, so its horizontal reach is negative while its vertical reach is positive.
Key Takeaways
Cos 5pi/6 equals $-\dfrac{\sqrt{3}}{2}$, about $-0.8660$ - negative because $\dfrac{5\pi}{6}$ lies in Quadrant II.
The reference angle is $\dfrac{\pi}{6}$, so the magnitude matches $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$; the quadrant supplies the minus sign.
In degrees, $\cos\dfrac{5\pi}{6} = \cos 150^\circ = -\dfrac{\sqrt{3}}{2}$, and the terminal point is $\left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)$.
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
Evaluate $4\cos\dfrac{5\pi}{6} + \sec\dfrac{5\pi}{6}$.
Without a calculator, decide the sign of $\cos\dfrac{7\pi}{6}$ and give its value.
Show that $\cos\dfrac{5\pi}{6} + \cos\dfrac{\pi}{6} = 0$ using the reference angle.
Want a live Bhanzu trainer to walk through Quadrant II signs and reference angles? Book a free demo class.
Read More
Cos 2pi/3 — another Quadrant II angle, where cosine is $-\dfrac{1}{2}$.
Cos 120 Degrees — the degree form of that Quadrant II partner.
Cosine function — how cosine behaves across all four quadrants.
Trigonometric ratios of specific angles — the full standard-angle set.
What is a radian — why $\dfrac{5\pi}{6}$ measures $150^\circ$.
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