Arcsin 1 — Value in Degrees and Radians

Arcsin 1 in Degrees and Radians

The value of arcsin 1 is $90°$, which in radians is $\dfrac{\pi}{2}$ (about $1.5708$ rad). It is the unique angle inside the inverse-sine range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ whose sine equals $1$.

Quick Answer:

• Result: arcsin 1 = 90° = π/2 radians

• Notation: arcsin 1 = sin⁻¹ 1 (the inverse function, NOT 1/sin)

• Method shown: read the angle whose sine is 1 from the unit circle / special-angle table

• Exact form: π/2 radians (= 90°)

• Approximate value: 1.5708 radians

Because sine peaks at $1$ exactly once in the inverse-sine range — at the top of the unit circle — there is only one angle that qualifies. That angle is $\dfrac{\pi}{2}$, the quarter turn.

Sin-Inverse Reference Table

These are the sin-inverse values readers look up alongside $\arcsin 1$. Each is shown in radians and degrees, with the unit-circle point that produces it.

Where Arcsin 1 Appears

The value $90°$ surfaces wherever a quantity hits its maximum. Is arcsin 1 the same as 90 degrees? Yes — and that matters in physics: a projectile launched at the angle whose normalized vertical component is $1$ travels straight up, the $90°$ extreme.

In signal processing, a sine wave reaches peak amplitude exactly when its phase angle is $\arcsin 1 = \pi/2$, which is why the quarter-period mark is the crest of every oscillation. Inverse sine also turns up in robotics, where a joint angle is recovered from a known sine ratio — and the $\pi/2$ output flags the arm at full extension.

What Arcsin Means

Arcsine answers the question "which angle has this sine?" Written $\arcsin x$ or $\sin^{-1} x$, it is the inverse of the sine function. Sine on its own is many-to-one — $\sin 90° = 1$, but so would $\sin 450°$ if we let the angle roam — so to define a single-valued inverse, mathematicians restrict sine to $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, where it climbs steadily from $-1$ to $1$. The inverse of that restricted sine is the arcsine, and its outputs live in that same interval.

Feed it $1$ and it returns the top of that range: $\dfrac{\pi}{2}$.

How to Find Arcsin 1

Method 1: Read it from the special-angle table

The sine table gives $\sin 90° = 1$. Arcsine reverses the lookup:

$$\arcsin 1 = 90° = \frac{\pi}{2}.$$

No other angle in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ has sine $1$, so the answer is unique.

Final answer: $\arcsin 1 = 90° = \dfrac{\pi}{2}$.

Method 2: Locate it on the unit circle

On the unit circle, the sine of an angle is the $y$-coordinate of its terminal point. The only point with $y = 1$ is $(0, 1)$, sitting at the top:

$$\arcsin 1 = \text{angle to } (0,1) = \frac{\pi}{2} = 90°.$$

Final answer: the same $\dfrac{\pi}{2} = 90°$, confirmed by the unit circle.

Examples of Arcsin 1

Example 1

Convert $\arcsin 1$ to radians.

From the special-angle table, $\arcsin 1 = 90°$. Multiply by $\dfrac{\pi}{180}$: $90 \times \dfrac{\pi}{180} = \dfrac{\pi}{2}$.

Final answer: $\dfrac{\pi}{2}$ radians.

Example 2

A student writes $\arcsin 1 = \dfrac{1}{\sin 1}$. Check the claim.

Wrong attempt. Reading $\sin^{-1}$ as a power, the student computes $(\sin 1)^{-1} = \dfrac{1}{\sin 1} \approx \dfrac{1}{0.0175} \approx 57.3$ (calculator in degrees) and reports $\arcsin 1 \approx 57.3$.

The break. Arcsine returns an angle whose sine is the input; the input here is $1$, and sine never exceeds $1$, so the angle must be the one where sine maxes out. A value near $57.3$ is not even close — and $\sin(57.3°) \approx 0.84$, not $1$.

Correct. The $-1$ in $\sin^{-1}$ marks the inverse function, not a reciprocal. The angle whose sine is $1$ is $90°$:

$$\arcsin 1 = \frac{\pi}{2} = 90°.$$

Final answer: $\arcsin 1 = \dfrac{\pi}{2} = 90°$, not $\dfrac{1}{\sin 1}$.

Example 3

Find $\arcsin(-1)$ in both units.

The point with $y = -1$ is $(0, -1)$, at the bottom of the unit circle, angle $-\dfrac{\pi}{2}$.

Final answer: $\arcsin(-1) = -\dfrac{\pi}{2} = -90°$.

Example 4

Evaluate $\arcsin 1 + \arccos 1$.

$\arcsin 1 = \dfrac{\pi}{2}$ and $\arccos 1 = 0$, so the sum is $\dfrac{\pi}{2}$. This matches the identity $\arcsin x + \arccos x = \dfrac{\pi}{2}$ at $x = 1$.

Final answer: $\dfrac{\pi}{2} = 90°$.

Example 5

A right triangle has its opposite side equal to its hypotenuse. What is the angle?

The angle's sine is $\dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{1} = 1$, so the angle is $\arcsin 1 = 90°$. (The triangle degenerates — the opposite side and hypotenuse coincide, which is the geometric limit of the right-triangle picture.)

Final answer: $90° = \dfrac{\pi}{2}$.

Where Sin-Inverse Trips Students Up

Mistake 1: Reading $\sin^{-1}$ as a reciprocal

Where it slips in: A reader sees $\sin^{-1} 1$ and computes $\dfrac{1}{\sin 1}$ instead of the inverse angle.

Don't do this: Treat the $-1$ as an exponent, the way it works on ordinary numbers.

The correct way: $\sin^{-1}$ is the inverse function, $\arcsin$. For the reciprocal of sine, write $\csc x$ or $\dfrac{1}{\sin x}$ — and that is the territory of the reciprocal identities, a separate idea. In

Mistake 2: Forgetting the input ceiling of 1

Where it slips in: A reader tries $\arcsin 2$ expecting an answer.

Don't do this: Feed arcsine a number outside $[-1, 1]$.

The correct way: Sine never exceeds $1$ in magnitude, so $\arcsin x$ is undefined as a real number for $|x| > 1$. The endpoint $\arcsin 1 = 90°$ is the largest legal output.

Mistake 3: Picking the wrong angle for sine 1

Where it slips in: A reader knows $\sin 90° = 1$ but also recalls $\sin 450° = 1$ and isn't sure which to report.

Don't do this: Return an angle outside the inverse-sine range.

The correct way: Arcsine outputs only angles in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, so $\dfrac{\pi}{2}$ is the answer and $\dfrac{5\pi}{2}$ is not — even though both have sine $1$.

What to Remember About Arcsin 1

• Arcsin 1 equals $90°$, or $\dfrac{\pi}{2}$ radians — the angle whose sine is exactly $1$.

• On the unit circle it is the point $(0, 1)$ at the top, where the $y$-coordinate hits its maximum.

• $\sin^{-1}$ means the inverse function, not the reciprocal $\dfrac{1}{\sin}$.

• Arcsine is undefined for inputs outside $[-1, 1]$, so $\arcsin 1$ is the largest possible output.

Practice These Three

1. Find $\arcsin\left(\dfrac{\sqrt{3}}{2}\right)$ in degrees and radians.

2. Evaluate $\arcsin 1 - \arcsin(-1)$.

3. A ramp rises so that the sine of its incline is $1$. What is the incline angle, and what does that mean physically?

If Problem 2 gives $\pi$, the two extremes of arcsine ($\pm \pi/2$) add to a straight angle.