What Does Sin 5pi/6 Mean?
The angle $\frac{5\pi}{6}$ is a rotation five-sixths of the way to $\pi$, landing in the second quadrant, the region past $90^\circ$ but before $180^\circ$. On the unit circle, sine is the $y$-coordinate of the point where the rotated radius meets the circle.
For $\frac{5\pi}{6}$ that point is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, so the sine is the $y$-value $\frac{1}{2}$. The size $\frac{1}{2}$ is inherited from the sin 30° computation; the sign stays positive.
Where Does Sin 5pi/6 Show Up?
Angles just short of half a turn appear whenever something rotates past vertical but not yet to the far side: a pendulum near the top of its swing, or a robotic arm angled at $150^\circ$, has a vertical reach proportional to $\sin\frac{5\pi}{6}$. Because the value is a clean $\frac{1}{2}$, the vertical component is exactly half the radius, a fact designers use for quick estimates.
The positive sign matters as much as the size. It tells you the point is still above the axis, which you can read directly from the unit circle.
Standard-Angle Reference Table
An angle in radians measures arc length on a unit circle instead of degrees. Here is $\frac{5\pi}{6}$ among the second-quadrant standard angles, with the corresponding sine values.
Angle (radians) | Angle (degrees) | Quadrant | $\sin\theta$ (exact) | $\sin\theta$ (decimal) |
|---|---|---|---|---|
$\dfrac{\pi}{2}$ | $90^\circ$ | Axis | $1$ | $1.0000$ |
$\dfrac{2\pi}{3}$ | $120^\circ$ | Second | $\dfrac{\sqrt{3}}{2}$ | $0.8660$ |
$\dfrac{3\pi}{4}$ | $135^\circ$ | Second | $\dfrac{\sqrt{2}}{2}$ | $0.7071$ |
$\dfrac{5\pi}{6}$ | $150^\circ$ | Second | $\dfrac{1}{2}$ | $0.5000$ |
$\pi$ | $180^\circ$ | Axis | $0$ | $0.0000$ |
Every sine value in the second quadrant is positive, because the $y$-coordinate stays above the $x$-axis even as the angle passes $90^\circ$. Written in the ratio form, the same angle also gives tan 5π/6, where the sign does turn negative.
How Do You Find The Exact Value Of Sin 5pi/6?
The reliable route uses the reference angle, then applies the quadrant sign separately.
Method 1: Reference angle plus quadrant sign.
The reference angle is the acute angle between the terminal radius and the $x$-axis. For a second-quadrant angle you subtract from $\pi$:
$$\pi - \frac{5\pi}{6} = \frac{\pi}{6}$$
The sine of the reference angle is $\sin\frac{\pi}{6} = \frac{1}{2}$. Sine is positive in the second quadrant, so the sign is unchanged:
$$\sin\frac{5\pi}{6} = +\sin\frac{\pi}{6} = \frac{1}{2}$$
Method 2: The unit circle point.
Rotate the radius $\frac{5\pi}{6}$ (that is, $150^\circ$) from the positive $x$-axis. The tip lands at $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, and the sine is the $y$-coordinate:
$$\sin\frac{5\pi}{6} = y\text{-coordinate} = \frac{1}{2}$$
Both methods return $\frac{1}{2}$, because the reference-angle rule is a shortcut for reading the same unit-circle point.
Examples Of Sin 5pi/6
Example 1
Evaluate $8\sin\frac{5\pi}{6}$.
$$8\sin\frac{5\pi}{6} = 8 \times \frac{1}{2} = 4$$
Example 2
A student says $\sin\frac{5\pi}{6} = -\frac{1}{2}$ because $150^\circ$ is "in the back half" of the circle.
Wrong attempt. The student assumes any angle past $90^\circ$ produces a negative sine.
That over-generalises the third-quadrant rule. In the second quadrant the $y$-coordinate is still above the axis, so the sine stays positive; a negative answer contradicts the point $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
Correct. The reference angle $\frac{\pi}{6}$ gives the size $\frac{1}{2}$, and second-quadrant sine is positive, so $\sin\frac{5\pi}{6} = \frac{1}{2}$. Only cosine and tangent turn negative here.
Example 3
Verify $\sin^2\frac{5\pi}{6} + \cos^2\frac{5\pi}{6} = 1$, given $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$.
$$\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$
The Pythagorean identity holds, and the negative cosine washes out under squaring.
Example 4
Use the supplement rule $\sin(\pi - x) = \sin x$ to find $\sin\frac{5\pi}{6}$.
Since $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$, the rule gives $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}$. This is why $\sin 150^\circ$ and $\sin 30^\circ$ share a value: they are supplementary angles.
Example 5
A wire is anchored so it makes an angle of $\frac{5\pi}{6}$ with the positive horizontal. If the wire is $10$ m long, find its vertical rise above the anchor.
$$\text{rise} = 10 \times \sin\frac{5\pi}{6} = 10 \times \frac{1}{2} = 5 \text{ m}$$
Where Students Trip Up On Sin 5pi/6
Mistake 1: Assuming past 90° means negative
Where it slips in: Applying the third-quadrant sign rule to a second-quadrant angle.
Don't do this: Writing $\sin\frac{5\pi}{6} = -\frac{1}{2}$ because the angle is obtuse.
The correct way: In the second quadrant only cosine and tangent are negative; sine stays positive. The learner who memorises "past $90^\circ$ is negative" without splitting the quadrants is the one who flips this sign.
Mistake 2: Subtracting from the wrong benchmark
Where it slips in: Computing the reference angle by subtracting $\frac{\pi}{2}$ or $\frac{5\pi}{6}$ from itself instead of from $\pi$.
Don't do this: Writing the reference angle as $\frac{5\pi}{6} - \frac{\pi}{2} = \frac{\pi}{3}$.
The correct way: In the second quadrant the reference angle is $\pi - \theta$, so $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. A reference angle must be acute; a wrong benchmark gives a wrong size.
Mistake 3: Confusing sin 5pi/6 with cos 5pi/6
Where it slips in: Reading the $x$-coordinate instead of the $y$-coordinate off the unit circle.
Don't do this: Writing $\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$, which is actually the cosine.
The correct way: Sine is the $y$-coordinate, cosine is the $x$-coordinate. At $\frac{5\pi}{6}$ the point is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, so sine is $\frac{1}{2}$ and cosine is $-\frac{\sqrt{3}}{2}$.
Key Takeaways
Sin 5pi/6 equals $\frac{1}{2}$ (that is $0.5$), an exact value because $\frac{5\pi}{6}$ is a standard angle.
The reference angle is $\frac{\pi}{6}$, giving the size $\frac{1}{2}$; the second-quadrant rule keeps the sign positive.
In degrees, $\frac{5\pi}{6} = 150^\circ$, and the unit-circle point is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
The most common slip is assuming any obtuse angle gives a negative sine; only cosine and tangent turn negative here.
To master quadrant signs with a teacher, explore Bhanzu's trigonometry tutor or high school math tutor, or browse math classes online.
Practice These Before Moving On
Evaluate $6\sin\frac{5\pi}{6} - 1$.
State the reference angle and the sign of sine for $\frac{7\pi}{6}$, then find $\sin\frac{7\pi}{6}$.
Use $\sin(\pi - x) = \sin x$ to show $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6}$.
Want a live Bhanzu trainer to walk through more sin 5pi/6 problems? Book a free demo class.
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