Arcsin

Idea

Between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ , the sine rises steadily from $-1$ to $1$ . It hits every value in that range exactly once, and never repeats.

That makes it the natural piece to invert. It is the widest piece with no repeats, and it is centred on zero.

Rule

y = \arcsin x \iff \sin y = x, \quad y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

D = [-1, 1]

R = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

The domain is where sine can land; the range is the window chosen to make the inverse single-valued. On the circle, the window is the green arc.

The unit circle

Try this. Try 0°, 90°, 180° and 270°. The horizontal projection is cosine; the vertical projection is sine. Track their signs between axes.

Angle (degrees)30

θ = 30° = 0.167π rad · cos θ = 0.866 · sin θ = 0.5 · arcsin window: -90 to 90 degrees: inside

How it is used

1. Arcsin is odd

The window straddles zero, so a negative input gives a negative angle.

\arcsin(-x) = -\arcsin x

Solve the positive case, then change the sign. Never treat a negative input separately. The Explore at the end of this lesson shows the symmetry on the graph.

2. The values worth knowing

$x$	$-1$	$-\dfrac{1}{2}$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arcsin x$	$-\dfrac{\pi}{2}$	$-\dfrac{\pi}{6}$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$
In degrees	$-90^\circ$	$-30^\circ$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$

3. Getting x out

Each function releases what the other traps. Apply it to both sides, then finish with ordinary algebra.

Trapped inside	Apply	Leaves
$\sin(\text{something}) = k$	$\arcsin$	$\text{something} = \arcsin k$
$\arcsin(\text{something}) = k$	$\sin$	$\text{something} = \sin k$

4. No special angle? Draw the triangle

Never reach for a calculator. Name the angle $\alpha$ , so $\sin\alpha$ is the given ratio.

$\sin\alpha = \dfrac{\text{opposite}}{\text{hypotenuse}}$

That is two sides. Pythagoras gives the third, and then any function of $\alpha$ can be read off. This method works for every inverse trig function, and the later lessons lean on it - learn it here.

5. A sum that needs no calculation

Two ratios whose squares add to $1$ are the legs of a right triangle with hypotenuse $1$ . The angles facing them are its two acute angles.

x^2 + y^2 = 1 \Rightarrow \arcsin x + \arcsin y = \frac{\pi}{2}

Check the condition first. If it holds, a hard-looking sum is free.

6. Two checks before you finish

Check	Why
A negative input gave a negative angle	arcsin never returns an obtuse angle
$\pm\dfrac{\pi}{2}$ is allowed	the range is closed, unlike arctan's

Worked example

Find the inverse of $f(x) = 5\sin(3x + 2) - 1$ . Write $y$ for $f(x)$ , then peel the expression apart from the outside in: $y = 5\sin(3x + 2) - 1$ . Move the constant and divide by the coefficient to leave the sine alone:

\frac{y + 1}{5} = \sin(3x + 2)

Now apply arcsin to both sides, which releases the whole bracket - not $x$ by itself:

\arcsin\left(\frac{y + 1}{5}\right) = 3x + 2

The rest is ordinary algebra: subtract $2$ , then divide by $3$ . Swapping the letters at the end gives

f^{-1}(x) = \frac{\arcsin\left(\frac{x + 1}{5}\right) - 2}{3}

Worked example - a ratio that is not special

Evaluate $\cos\left(\arcsin\dfrac{3}{5}\right)$ .

Name the angle: let $\alpha = \arcsin\dfrac{3}{5}$ , which says $\sin\alpha = \dfrac{3}{5}$ . So the triangle has opposite $3$ and hypotenuse $5$ , and Pythagoras supplies the third side: $\sqrt{25 - 9} = 4$ . The triangle is complete, so the answer can be read straight off it:

\cos\alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}

The same triangle also gives $\tan\alpha = \dfrac{3}{4}$ , and any other function of that angle. That is why you draw the triangle instead of solving for one value at a time.

A right triangle

Try this. Compare the labeled lengths with sine, cosine and tangent. If size is adjustable, scale the triangle: lengths change together, but the ratios stay fixed.

θ = 36.87° · sin θ = 0.6 · cos θ = 0.8 · tan θ = 0.75

Explore

Drag the probe. The two marks sit at $x$ and $-x$ , and their readings are always the same distance either side of zero - that is what odd means.

Ride $x$ out to $1$ : the angle reaches $90$ degrees exactly, because arcsin's range is closed. Sine never goes past $1$ , so past the edge arcsin has nothing to say.

Inverse trigonometric functions

Try this. Move the probe from positive x to negative x. Compare the two outputs in radians and the indicated symmetry or sum.

Probe x0.5

arcsin(0.5) = 0.524 rad · arcsin(-0.5) = -0.524 rad