Inverse Trigonometric Functions

Inverse trigonometric functions was worth 4 marks on NESA's sample paper for the 2024 syllabus, usually as a multiple choice question on a domain, range or exact value plus a short Section II part. Their derivatives and integrals are in Further calculus skills (ME12-CS), where they are worth much more, so the definitions here matter.

• Definitions: why $\sin$, $\cos$ and $\tan$ need restricted domains to have inverses; the definitions of ${\sin }^{-1}x$, ${\cos }^{-1}x$ and ${\tan }^{-1}x$ with their domains and ranges.
• Exact values and simplifying using the definitions.
• Graphs: reflecting the restricted functions in $y=x$, odd and even properties, and transformations (reflections, translations, dilations) including domain, range and asymptotes.
• Properties: prove and use ${\sin }^{-1}(-x)=-{\sin }^{-1}x$, ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$, ${\tan }^{-1}(-x)=-{\tan }^{-1}x$ and ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$.
• Composites: graph $\sin ({\sin }^{-1}x)$, ${\sin }^{-1}(\sin x)$ and the cosine and tangent versions, and know where each simplifies to $x$.

All angles are in radians.

A function has an inverse function only if it is one-to-one, that is, its graph passes the horizontal line test. The graphs of $\sin x$, $\cos x$ and $\tan x$ fail it, so each is restricted to one piece that takes every value exactly once.

So ${\sin }^{-1}x$ means "the angle between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ whose sine is $x$". These are also written $\arcsin x$, $\arccos x$ and $\arctan x$. Note that ${\sin }^{-1}x$ is not $\frac{1}{\sin x}$.

The ranges are what make exact-value questions work. A negative input gives a negative angle for ${\sin }^{-1}$ and ${\tan }^{-1}$, but a second-quadrant angle for ${\cos }^{-1}$.

Worked example 1: exact values

Find the exact values of (a) ${\cos }^{-1}\left(-\frac{1}{2}\right)$, (b) $\tan \left({\sin }^{-1}\left(-\frac{8}{17}\right)\right)$ and (c) ${\tan }^{-1}\left(\tan \frac{3\pi }{4}\right)$.

(a) The angle in $[0,\pi ]$ with cosine $-\frac{1}{2}$ is $\frac{2\pi }{3}$.

(b) Let $\alpha ={\sin }^{-1}\left(-\frac{8}{17}\right)$. Then $\sin \alpha =-\frac{8}{17}$ and $-\frac{\pi }{2}\le \alpha \le \frac{\pi }{2}$, so $\alpha$ is in the fourth quadrant where cosine is positive. A right triangle with opposite 8 and hypotenuse 17 has adjacent side $\sqrt{17^2-8^2}=15$, so $\cos \alpha =\frac{15}{17}$ and

(c) $\tan \frac{3\pi }{4}=-1$, and the angle in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ with tangent $-1$ is $-\frac{\pi }{4}$. The answer is not $\frac{3\pi }{4}$, because $\frac{3\pi }{4}$ is outside the range of ${\tan }^{-1}$.

Each inverse graph is the reflection of the restricted trig graph in $y=x$.

• $y={\sin }^{-1}x$ rises from $\left(-1,-\frac{\pi }{2}\right)$ through the origin to $\left(1,\frac{\pi }{2}\right)$. It is odd.
• $y={\cos }^{-1}x$ falls from $(-1,\pi )$ through $\left(0,\frac{\pi }{2}\right)$ to $(1,0)$. It is neither odd nor even, but it has point symmetry about $\left(0,\frac{\pi }{2}\right)$.
• $y={\tan }^{-1}x$ rises through the origin with horizontal asymptotes $y=\pm \frac{\pi }{2}$. It is odd.

For a transformed function such as $y=a{\sin }^{-1}(bx+c)+d$, find the domain by solving $-1\le bx+c\le 1$, and the range by applying $a$ and $d$ to the basic range (a negative $a$ flips the ends). For ${\tan }^{-1}$, the asymptotes move with the range.

Worked example 2: domain, range and asymptotes

State the domain and range of (a) $y=2{\sin }^{-1}(1-x)$ and (b) $y=4{\tan }^{-1}(2x)-\pi$, including any asymptotes.

(a) Need $-1\le 1-x\le 1$, which gives $0\le x\le 2$. The basic range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is doubled: $-\pi \le y\le \pi$. Because of the $-x$ inside, the graph decreases from $(0,\pi )$ to $(2,-\pi )$.

(b) The domain is all real $x$. Multiply the range $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ by 4 and subtract $\pi$:

with horizontal asymptotes $y=-3\pi$ and $y=\pi$.

The four properties

• ${\sin }^{-1}(-x)=-{\sin }^{-1}x$ and ${\tan }^{-1}(-x)=-{\tan }^{-1}x$ (odd functions).
• ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$.
• ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ for $-1\le x\le 1$.

To prove ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$: let $\theta ={\cos }^{-1}x$, so $\cos \theta =x$ and $0\le \theta \le \pi$. Then $\cos (\pi -\theta )=-\cos \theta =-x$, and $0\le \pi -\theta \le \pi$, so $\pi -\theta$ is in the range of ${\cos }^{-1}$. Hence ${\cos }^{-1}(-x)=\pi -\theta$. The range check is the step students leave out.

The proof of ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ is similar: let $\alpha ={\sin }^{-1}x$, then $\cos \left(\frac{\pi }{2}-\alpha \right)=\sin \alpha =x$, and $\frac{\pi }{2}-\alpha$ lies in $[0,\pi ]$.

Composites

• $\sin ({\sin }^{-1}x)=x$ for $-1\le x\le 1$ only (outside that, it is undefined). The same holds for $\cos$; $\tan ({\tan }^{-1}x)=x$ for all $x$.
• ${\sin }^{-1}(\sin x)=x$ only for $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$. It is defined for all $x$ but outside that interval its graph is a zigzag of straight line segments between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$.
• ${\cos }^{-1}(\cos x)=x$ only for $0\le x\le \pi$; ${\tan }^{-1}(\tan x)=x$ only for $-\frac{\pi }{2}<x<\frac{\pi }{2}$.

Worked example 3: using a property to solve an equation

Solve ${\sin }^{-1}x=2{\cos }^{-1}x$.

Replace ${\cos }^{-1}x$ with $\frac{\pi }{2}-{\sin }^{-1}x$:

Check: ${\sin }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{3}$ and $2{\cos }^{-1}\frac{\sqrt{3}}{2}=2\times \frac{\pi }{6}=\frac{\pi }{3}$. Also $\frac{\pi }{3}$ is inside the range of ${\sin }^{-1}$, so the solution is valid.

• Giving an answer outside the range, for example ${\sin }^{-1}\left(\sin \frac{4\pi }{3}\right)=\frac{4\pi }{3}$. Always check the range table.
• Writing ${\cos }^{-1}(-x)=-{\cos }^{-1}x$. Inverse cosine is not odd.
• Treating ${\sin }^{-1}x$ as $\frac{1}{\sin x}$.
• Finding the range of $y=a{\sin }^{-1}(\dots )$ but forgetting to swap the ends when $a<0$.
• Using degrees, or a calculator in degree mode.
• Drawing $y={\tan }^{-1}x$ crossing its asymptotes, or giving it vertical asymptotes.

• For exact values, draw a right-angled triangle and decide the quadrant from the range of the inverse function before you write the sign.
• Sketch questions: label endpoints with exact coordinates (for ${\sin }^{-1}$ and ${\cos }^{-1}$) or the asymptotes with their equations (for ${\tan }^{-1}$), and the intercepts.
• When proving a property, the mark often depends on stating that the new angle lies in the correct range.
• Domain and range questions are popular multiple choice items. Work them out directly rather than testing options.
• If a question says "hence", use the identity just proved, as in worked example 3.

• Further work with functions (ME11-FN): inverse functions, the horizontal line test and reflection in $y=x$.
• Further calculus skills (ME12-CS): derivatives of ${\sin }^{-1}$, ${\cos }^{-1}$, ${\tan }^{-1}$ and the integrals that give them.
• Further applications of calculus (ME12-CA): solving differential equations such as $\frac{dy}{dx}=1+y^2$ produces ${\tan }^{-1}$.
• Introduction to vectors (ME12-VE): angles between vectors and directions of motion use ${\cos }^{-1}$ and ${\tan }^{-1}$.
• Advanced, Graph Transformations (MA11-GT): the same rules for dilations, translations and reflections.