Compositions with Trig Functions

Two Ways to Combine Them

Once you have both a trig function and its inverse, you can chain them together. The order matters, and the two orders behave very differently.

• trig of inverse: like $\sin(\arcsin x)$
• inverse of trig: like $\arcsin(\sin\theta)$

One of these always cancels cleanly. The other has a trap in it.

The Clean Direction: trig of inverse

Feed a value into an inverse, then into the matching trig function, and you get the value straight back:

$\sin(\arcsin x) = x \qquad \cos(\arccos x) = x \qquad \tan(\arctan x) = x$

The inverse hands the outer function exactly the angle it needs, so nothing is lost.

The only rule is that $x$ must be a legal input for the inverse. $\sin(\arcsin 2)$ is meaningless, because $\arcsin 2$ does not exist ($2$ is outside $[-1, 1]$).

The Trap: inverse of trig

Now the other order. You might expect $\arcsin(\sin\theta) = \theta$ always. It does not.

The gray point is the real angle $\theta$, sweeping all the way around. The green point is $\arcsin(\sin\theta)$, which can only ever answer with an angle in the right half of the circle. When $\theta$ wanders into the left half, the answer folds back to the angle there with the same height.

$\arcsin(\sin\theta) = \theta$ only when $\theta$ is already in $[-90°, 90°]$.

Seeing the Fold in Numbers

Take $\theta = 150°$.

• $\sin 150° = \frac{1}{2}$
• $\arcsin\frac{1}{2} = 30°$

So $\arcsin(\sin 150°) = 30°$, not $150°$.

Both $30°$ and $150°$ have a sine of $\frac{1}{2}$, but only $30°$ lives in the principal range. The inverse always returns the principal angle that shares the value, which equals $\theta$ only when $\theta$ was already in range.

Mixed Compositions: a Right Triangle

The most useful case mixes different functions, like $\sin(\arccos x)$. There is no direct cancel here, but a right triangle turns it into simple algebra.

Let $\theta = \arccos x$, so $\cos\theta = x$. Since cosine is $\frac{\text{adjacent}}{\text{hypotenuse}}$, build a triangle with adjacent $= x$ and hypotenuse $= 1$. The Pythagorean theorem fills in the last side:

$\text{opposite} = \sqrt{1 - x^2}$

Now just read off the sine:

$\sin(\arccos x) = \sqrt{1 - x^2}$

The Common Ones

The same triangle trick gives a whole family of clean algebraic results:

These matter because they replace a tangled trig expression with a plain formula. That swap is used constantly in calculus.

Example

Find $\cos(\arcsin\frac{3}{5})$.

Let $\theta = \arcsin\frac{3}{5}$, so $\sin\theta = \frac{3}{5} = \frac{\text{opposite}}{\text{hypotenuse}}$.

That is the famous $3$-$4$-$5$ triangle: opposite $3$, hypotenuse $5$, so adjacent $= 4$.

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$

So $\cos(\arcsin\frac{3}{5}) = \frac{4}{5}$.