Sin 45 Degrees : Value √2/2 Explained

What Does Sin 45 Degrees Mean?

Sine is one of the three trigonometric ratios, in a right triangle, the sine of an angle is the side opposite it divided by the hypotenuse. So $\sin 45^\circ$ asks: in a right triangle with a $45^\circ$ angle, what fraction of the hypotenuse is the opposite side?

On the unit circle, a circle of radius $1$ centred at the origin, sine is the $y$-coordinate of the point where the angle's radius meets the circle. At $45^\circ$ that point is $\left(\dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{2}}{2}\right)$, so the $y$-coordinate, and the sine, is $\dfrac{\sqrt{2}}{2}$.

Because the point sits on the line $y = x$, its two coordinates are equal, which is the geometric reason $\sin 45^\circ = \cos 45^\circ$.

Where Does Sin 45 Degrees Show Up?

A $45^\circ$ launch angle gives a projectile its maximum range on level ground, and both its horizontal and vertical components scale with $\sin 45^\circ = \dfrac{\sqrt{2}}{2}$, equal, which is exactly why $45^\circ$ is the sweet spot. The same value sets the diagonal of a square: a square of side $1$ has a diagonal of $\sqrt{2}$, and the diagonal makes a $45^\circ$ angle with each side.

Screen and camera diagonals, isometric drawings, and banked $45^\circ$ turns all lean on this number. It is the most-used non-trivial sine after $30^\circ$ and $60^\circ$.

Standard-Angle Reference Table

Forty-five degrees is one of the handful of angles whose sine has a clean exact form. Here are the standard first-quadrant angles in both degrees and radians.

Read the column top to bottom and sine climbs from $0$ up to $1$ as the angle opens. At $45^\circ$ it passes through $\dfrac{\sqrt{2}}{2}$, the exact midpoint of the circle's quarter-turn, and the only standard angle where $\sin\theta = \cos\theta$.

How Do You Find The Exact Value Of Sin 45 Degrees?

Two clean routes both give $\dfrac{\sqrt{2}}{2}$.

Method 1: The 45-45-90 triangle.

Take an isosceles right triangle with both legs equal to $1$. Its two base angles are each $45^\circ$, and by the Pythagorean theorem the hypotenuse is

$$\sqrt{1^2 + 1^2} = \sqrt{2}$$

Now apply the definition, with the side opposite the $45^\circ$ angle equal to $1$:

$$\sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

The last step rationalises the denominator by multiplying top and bottom by $\sqrt{2}$.

Method 2: The unit circle.

Rotate the radius $45^\circ$ above the positive $x$-axis. Because $45^\circ$ splits the first quadrant evenly, the point lands on the line $y = x$ at $\left(\dfrac{\sqrt{2}}{2},\ \dfrac{\sqrt{2}}{2}\right)$.

$$\sin 45^\circ = y\text{-coordinate} = \frac{\sqrt{2}}{2}$$

The radian name for this angle is $\dfrac{\pi}{4}$, so $\sin\dfrac{\pi}{4} = \sin 45^\circ = \dfrac{\sqrt{2}}{2}$. The radian-first walkthrough lives at sin π/4.

Examples Of Sin 45 Degrees

Example 1

Evaluate $8\sin 45^\circ$.

$$8\sin 45^\circ = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \approx 5.657$$

Example 2

Find $\sin 45^\circ$ given that $\sin 45^\circ = \cos\theta$ for an acute angle $\theta$. A student writes $\theta = 45^\circ$ and stops, then a second student says "sine and cosine are always different, so $\theta$ can't be $45^\circ$."

Wrong attempt. The second student assumes sine and cosine can never coincide.

That breaks at exactly this angle: $45^\circ$ is the one place where $\sin\theta = \cos\theta$, both equal to $\dfrac{\sqrt{2}}{2}$. So $\theta = 45^\circ$ is right after all.

Correct. By the cofunction rule $\cos\theta = \sin(90^\circ - \theta)$, and $\sin 45^\circ = \cos 45^\circ$, so $\theta = 45^\circ$. The first student had it; the "always different" reasoning was the false step.

Example 3

A right triangle has a hypotenuse of $10$ cm and a $45^\circ$ angle. Find the side opposite the $45^\circ$ angle.

$$\sin 45^\circ = \frac{\text{opposite}}{10} \implies \text{opposite} = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07 \text{ cm}$$

Example 4

Verify $\sin^2 45^\circ + \cos^2 45^\circ = 1$.

$$\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1$$

The Pythagorean identity holds, and here both terms are equal because sine and cosine match at $45^\circ$.

Example 5

A ball is thrown at $45^\circ$ with speed $20$ m/s. Find the vertical component of its velocity.

$$v_y = 20 \times \sin 45^\circ = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2} \approx 14.14 \text{ m/s}$$

The horizontal component uses $\cos 45^\circ$ and comes out identical, which is why a $45^\circ$ throw balances height and distance.

Where Students Trip Up On Sin 45 Degrees

Mistake 1: Reporting 1/√2 when the standard form is asked

Where it slips in: Reading the value straight off the triangle ratio and stopping at $\dfrac{1}{\sqrt{2}}$.

Don't do this: Writing $\sin 45^\circ = \dfrac{1}{\sqrt{2}}$ on a problem that asks for the standard rationalised form.

The correct way: Rationalise: $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. Both are the same number, but $\dfrac{\sqrt{2}}{2}$ is the conventional exam form.

Mistake 2: Mixing up sin 45° with sin 30° or sin 60°

Where it slips in: Attaching the wrong radical to the wrong angle under time pressure.

Don't do this: Writing $\sin 45^\circ = \dfrac{\sqrt{3}}{2}$. That is $\sin 60^\circ$.

The correct way: Anchor on the pattern $\dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}$ for $\sin 30^\circ, \sin 45^\circ, \sin 60^\circ$. The memorizer who stores the three radicals in order rarely swaps them.

Mistake 3: Forgetting the calculator's angle mode

Where it slips in: A calculator in radian mode returns $\sin(45) \approx 0.851$ instead of $0.7071$.

Don't do this: Trusting the screen without checking whether it is set to degrees.

The correct way: Confirm degree mode before entering $\sin(45)$; a reading that is not near $0.71$ signals a mode error.

Key Takeaways

• Sin 45 degrees equals $\dfrac{\sqrt{2}}{2}$, about $0.7071$, an exact value because $45^\circ$ is a standard angle.

• The $45$-$45$-$90$ triangle gives it as $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$; the unit circle gives it as the $y$-coordinate on the line $y = x$.

• It is the only first-quadrant angle where $\sin\theta = \cos\theta$.

• The common slips are leaving the answer as $\dfrac{1}{\sqrt{2}}$ when the rationalised form is asked, and swapping it with $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$.

To drill the special angles with a teacher, explore Bhanzu's trigonometry tutor or its online math classes. The reference-angle cousin of this value in Quadrant II is sin 3π/4.

Practice These Before Moving On

1. Evaluate $2\sin 45^\circ + \cos 45^\circ$.

2. A square has side $6$ cm. Use $\sin 45^\circ$ to find the length of its diagonal.

3. Show that $\sin 45^\circ \times \cos 45^\circ = \dfrac{1}{2}$.

Want a live Bhanzu trainer to walk through more special-angle problems? Book a free demo class, online with an expert, anywhere.

For the geometry behind the value, see the reference on the special right triangle.

Read More

• 45 degree angle, the angle itself, from a geometry view.

• Unit circle, where every special-angle sine is read.

• Trigonometric table, sine, cosine, tangent at the standard angles.

• Trigonometric ratios of specific angles, the full special-angle set.

• Sin cos tan, the three ratios and how they connect.