What Does Sin 3pi/4 Mean?
Sine, on the unit circle, is the $y$-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 I to IV counter-clockwise from the top right.
The reference angle is the acute angle between the terminal side and the $x$-axis. For $\dfrac{3\pi}{4}$, the terminal side sits in Quadrant II, and the acute gap to the negative $x$-axis is $\pi - \dfrac{3\pi}{4} = \dfrac{\pi}{4}$.
That reference angle is why $\sin\dfrac{3\pi}{4}$ shares its size with $\sin\dfrac{\pi}{4}$. The two are reflections across the vertical axis, and reflection leaves the height unchanged.
Where Does Sin 3pi/4 Show Up?
A projectile launched at $135^\circ$ from the positive $x$-axis, or a vector pointing up-and-to-the-left, has a vertical component proportional to $\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$. The same value appears whenever a $45^\circ$ tilt is reflected into the second quadrant, the diagonal of a square, rotated.
It is also the standard first example of a Quadrant II angle in a trig course, because it shows the split cleanly: the size comes from the reference angle, the sign comes from the quadrant.
Standard-Angle Reference Table
$\dfrac{3\pi}{4}$ is one of the "$45^\circ$ family" of angles, the ones whose reference angle is $\dfrac{\pi}{4}$. Here are the sine values around the circle for that family, in radians and degrees.
Angle (radians) | Angle (degrees) | Quadrant | $\sin\theta$ |
|---|---|---|---|
$\dfrac{\pi}{4}$ | $45^\circ$ | I | $\dfrac{\sqrt{2}}{2}$ |
$\dfrac{3\pi}{4}$ | $135^\circ$ | II | $\dfrac{\sqrt{2}}{2}$ |
$\dfrac{5\pi}{4}$ | $225^\circ$ | III | $-\dfrac{\sqrt{2}}{2}$ |
$\dfrac{7\pi}{4}$ | $315^\circ$ | IV | $-\dfrac{\sqrt{2}}{2}$ |
The size of the sine is always $\dfrac{\sqrt{2}}{2}$ across the family; only the sign changes with the quadrant. Sine is positive above the horizontal axis (Quadrants I and II) and negative below it.
How Do You Find The Exact Value Of Sin 3pi/4?
Two routes give $\dfrac{\sqrt{2}}{2}$.
Method 1: Reference angle plus quadrant sign.
First find the reference angle. In Quadrant II it is $\pi$ minus the angle:
$$\pi - \frac{3\pi}{4} = \frac{\pi}{4}$$
The sine of the reference angle is $\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$. Quadrant II is above the axis, so sine is positive there. Therefore:
$$\sin\frac{3\pi}{4} = +\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$
This shares its exact value with sin 45°, the degree name for the reference angle, and with sin π/4.
Method 2: Read the unit circle.
Convert to degrees: $\dfrac{3\pi}{4} = 135^\circ$. Rotating $135^\circ$ lands the radius in Quadrant II at the point $\left(-\dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{2}}{2}\right)$.
$$\sin\frac{3\pi}{4} = y\text{-coordinate} = \frac{\sqrt{2}}{2}$$
The $x$-coordinate is negative (left of centre) but the $y$-coordinate is positive (above centre), which is exactly why cosine is negative here while sine stays positive.
Examples Of Sin 3pi/4
Example 1
Evaluate $6\sin\dfrac{3\pi}{4}$.
$$6\sin\frac{3\pi}{4} = 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2} \approx 4.243$$
Example 2
Find $\sin\dfrac{3\pi}{4}$. A student says "$\dfrac{3\pi}{4}$ is past $\dfrac{\pi}{2}$, so the sine must be negative."
Wrong attempt. Assuming that going past $\dfrac{\pi}{2}$ flips the sign of sine.
That breaks against the quadrant rule: $\dfrac{3\pi}{4}$ is in Quadrant II, which is still above the $x$-axis, so the $y$-coordinate, and the sine, is positive. The sign flips only when the point drops below the axis, in Quadrants III and IV.
Correct. Reference angle $\dfrac{\pi}{4}$ gives size $\dfrac{\sqrt{2}}{2}$; Quadrant II keeps sine positive. So $\sin\dfrac{3\pi}{4} = +\dfrac{\sqrt{2}}{2}$.
Example 3
Show that $\sin\dfrac{3\pi}{4} = \sin\dfrac{\pi}{4}$.
Both equal $\dfrac{\sqrt{2}}{2}$. By the identity $\sin(\pi - x) = \sin x$ with $x = \dfrac{\pi}{4}$:
$$\sin\frac{3\pi}{4} = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$
Example 4
Verify $\sin^2\dfrac{3\pi}{4} + \cos^2\dfrac{3\pi}{4} = 1$.
With $\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$ and $\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}$:
$$\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1$$
The Pythagorean identity holds even though cosine is negative, squaring erases the sign.
Example 5
A right-triangle problem gives $\sin\theta = \dfrac{\sqrt{2}}{2}$ with $\theta$ obtuse. Find $\theta$ in radians.
Two angles have sine $\dfrac{\sqrt{2}}{2}$: $\dfrac{\pi}{4}$ (acute) and $\dfrac{3\pi}{4}$ (obtuse). Since $\theta$ is obtuse, $\theta = \dfrac{3\pi}{4}$.
Where Students Trip Up On Sin 3pi/4
Mistake 1: Making sine negative in Quadrant II
Where it slips in: Remembering "second quadrant means some functions go negative" and applying it to the wrong one.
Don't do this: Writing $\sin\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}$. That is $\cos\dfrac{3\pi}{4}$; cosine is the negative one in Quadrant II, not sine.
The correct way: In Quadrant II, sine (the $y$-coordinate) stays positive and cosine (the $x$-coordinate) turns negative. So $\sin\dfrac{3\pi}{4} = +\dfrac{\sqrt{2}}{2}$.
Mistake 2: Using the wrong reference angle
Where it slips in: Subtracting from the wrong axis and getting $\dfrac{3\pi}{4}$ itself, or $\dfrac{\pi}{2}$.
Don't do this: Treating $\dfrac{3\pi}{4}$ as its own reference angle.
The correct way: In Quadrant II the reference angle is $\pi - \theta$. Here $\pi - \dfrac{3\pi}{4} = \dfrac{\pi}{4}$, the acute gap to the negative $x$-axis.
Mistake 3: Leaving √2/2 as an un-rationalised 1/√2 when the standard form is asked
Where it slips in: Reading the value off a calculator or a memorised $\dfrac{1}{\sqrt{2}}$ and stopping there.
Don't do this: Reporting $\dfrac{1}{\sqrt{2}}$ on a problem that asks for the standard rationalised form.
The correct way: Rationalise the denominator: $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. Both are correct numbers, but $\dfrac{\sqrt{2}}{2}$ is the conventional exam form.
Key Takeaways
Sin 3pi/4 equals $\dfrac{\sqrt{2}}{2}$, about $0.7071$, an exact value from the $45^\circ$ family.
The size comes from the reference angle $\dfrac{\pi}{4}$; the positive sign comes from Quadrant II.
In degrees, $\sin\dfrac{3\pi}{4} = \sin 135^\circ$, and it equals $\sin\dfrac{\pi}{4}$ by the identity $\sin(\pi - x) = \sin x$.
The common slip is making sine negative in Quadrant II, that is cosine's job, not sine's.
To master reference angles across the circle with a teacher, explore Bhanzu's trigonometry tutor or its online math classes. The same reference angle drives its degree twin, sin 45°, linked above.
Practice These Before Moving On
Evaluate $4\sin\dfrac{3\pi}{4} - 2\cos\dfrac{3\pi}{4}$.
Find every angle in $[0, 2\pi)$ whose sine is $\dfrac{\sqrt{2}}{2}$.
A vector makes a $135^\circ$ angle with the positive $x$-axis. Use $\sin\dfrac{3\pi}{4}$ to find its vertical component if its length is $8$.
Want a live Bhanzu trainer to walk through more Quadrant II problems? Book a free demo class, online with an expert, anywhere.
For the geometry behind the $45^\circ$ value, see the reference on the special right triangle.
Read More
Sin 5π/4, the Quadrant III member of the same family.
Sin 3π, a large radian angle reduced on the circle.
Quadrant, the four regions and their sign rules.
Trigonometric ratios of specific angles, the full special-angle set.
Trigonometric table, sine, cosine, tangent at the standard angles.
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