Inverse trigonometric functionsEdexcel International A Level Maths: Revision notes

A function has an inverse only if it is one-to-one. $\sin x$, $\cos x$ and $\tan x$ each repeat their values, so we restrict each to an interval where it is one-to-one. The inverse then undoes the original: if $y=\arcsin x$ then $x=\sin y$. The graph of an inverse is the reflection of the restricted original in the line $y=x$, so the domain and range swap. The values returned are called principal values.

$\arcsin x$ (or $\sin^{-1}x$) is the inverse of $\sin x$ restricted to $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$ (that is, $-90^{\circ}\le x\le90^{\circ}$).

• Domain: $-1\le x\le1$.
• Range: $-\frac{\pi}{2}\le y\le\frac{\pi}{2}$. Its graph is increasing and passes through the origin, ending at $\left(1,\frac{\pi}{2}\right)$ and $\left(-1,-\frac{\pi}{2}\right)$. Examples: $\arcsin\frac12=\frac{\pi}{6}$ and $\arcsin(-1)=-\frac{\pi}{2}$.

$\arccos x$ is the inverse of $\cos x$ restricted to $0\le x\le\pi$ ($0^{\circ}$ to $180^{\circ}$). Domain $-1\le x\le1$; range $0\le y\le\pi$. Its graph is decreasing, from $(-1,\pi)$ through $\left(0,\frac{\pi}{2}\right)$ to $(1,0)$. Example: $\arccos\left(-\frac12\right)=\frac{2\pi}{3}$. $\arctan x$ is the inverse of $\tan x$ restricted to $-\frac{\pi}{2}<x<\frac{\pi}{2}$. Domain: all real $x$; range $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Its graph is increasing through the origin, with horizontal asymptotes $y=\pm\frac{\pi}{2}$. Examples: $\arctan1=\frac{\pi}{4}$, $\arctan\sqrt3=\frac{\pi}{3}$ and $\arctan\left(-\sqrt3\right)=-\frac{\pi}{3}$.

Functions and inverses cancel only on the right interval:

• $\sin(\arcsin x)=x$ for $-1\le x\le1$, and $\cos(\arccos x)=x$ for $-1\le x\le1$;
• $\tan(\arctan x)=x$ for all real $x$;
• $\arcsin(\sin x)=x$ only when $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$, and $\arccos(\cos x)=x$ only when $0\le x\le\pi$. For example $\arcsin\left(\sin150^{\circ}\right)=\arcsin\frac12=30^{\circ}$, not $150^{\circ}$. For a value like $\cos(\arcsin\frac45)$, draw a right-angled triangle or use $\sin^2+\cos^2=1$: the angle is acute, so the answer is $\frac35$.

A calculator gives only the principal value. To find all solutions in a given interval, use the symmetry or period of the original function. Example. Solve $\tan x=-3$ for $-\pi<x<\pi$. $x=\arctan(-3)=-1.249$, and adding the period $\pi$ gives $x=1.893$. So $x=-1.25$ or $1.89$ (3 s.f.). Inverse functions also find angles in context. For a mast of height $12$ m seen from $d$ m away, $\theta=\arctan\frac{12}{d}$; at $d=5$, $\theta=67.4^{\circ}$. Remember that $\arcsin k$ and $\arccos k$ exist only for $-1\le k\le1$.