Sin 3pi/4 : Exact Value √2/2 and How to Find It

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.

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

1. Evaluate $4\sin\dfrac{3\pi}{4} - 2\cos\dfrac{3\pi}{4}$.

2. Find every angle in $[0, 2\pi)$ whose sine is $\dfrac{\sqrt{2}}{2}$.

3. 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.