Trigonometric and inverse trigonometric functions

Unit 5 Trigonometric and inverse trigonometric functions

Key unit competence: Use trigonometric and inverse trigonometric functions, their limits and derivatives in solving related problems.

Key ideas

• Domains: $\sin$, $\cos$: $\mathbb{R}$; $\tan$: $x\ne \frac{\pi }{2}+k\pi$; $\cot$: $x\ne k\pi$; $\sec$: $x\ne \frac{\pi }{2}+k\pi$; $\csc$: $x\ne k\pi$. Ranges: $\sin ,\cos \in [-1,1]$; $\tan ,\cot \in \mathbb{R}$; $\sec ,\csc \in ]-\infty ,-1]\cup [1,+\infty [$.
• Inverse functions (principal values): $\arcsin :[-1,1]\to \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$; $\arccos :[-1,1]\to [0,\pi ]$; $\arctan :\mathbb{R}\to \left]-\frac{\pi }{2},\frac{\pi }{2}\right[$.
• Key limit: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ (and ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$).
• Derivatives: $(\sin x{)}^{\prime }=\cos x$; $(\cos x{)}^{\prime }=-\sin x$; $(\tan x{)}^{\prime }={\sec }^{2}x=1+{\tan }^{2}x$; $(\cot x{)}^{\prime }=-{\csc }^{2}x$; $(\sec x{)}^{\prime }=\sec x\tan x$; $(\csc x{)}^{\prime }=-\csc x\cot x$.
• Inverse derivatives: $(\arcsin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}$; $(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}$; $(\arctan x{)}^{\prime }=\frac{1}{1+x^2}$; $(\mathrm{arccot}x{)}^{\prime }=-\frac{1}{1+x^2}$.
• Chain rule: $(\sin u{)}^{\prime }=u^{\prime }\cos u$, $(\arctan u{)}^{\prime }=\frac{u^{\prime }}{1+u^2}$, etc. Successive derivatives of $\sin$ repeat every 4: ${\sin }^{(n)}x=\sin \left(x+\frac{n\pi }{2}\right)$.

Formulas to know

Common misconceptions

• "${\sin }^{-1}x=\frac{1}{\sin x}$." No: ${\sin }^{-1}x$ means $\arcsin x$; $\frac{1}{\sin x}=\csc x$.
• "$\arcsin \left(\sin \frac{5\pi }{6}\right)=\frac{5\pi }{6}$." No: the answer must lie in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$: it is $\frac{\pi }{6}$.
• "$(\cos x{)}^{\prime }=\sin x$." No: $(\cos x{)}^{\prime }=-\sin x$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find ${\lim }_{x\to 0}\frac{\sin 4x}{\sin 2x}$.

   Show answer

   $\frac{4x}{2x}\to 2$ (or $\frac{2\sin 2x\cos 2x}{\sin 2x}=2\cos 2x\to 2$).

   Common mistake: Cancelling the "sin".

2. State the domain and range of $y=\arctan x$ and sketch its horizontal asymptotes.

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   Domain $\mathbb{R}$, range $\left]-\frac{\pi }{2},\frac{\pi }{2}\right[$; asymptotes $y=\pm \frac{\pi }{2}$.

   Common mistake: Including $\pm \frac{\pi }{2}$ in the range.

3. Differentiate $y={\tan }^{3}x$.

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   $3{\tan }^{2}x{\sec }^{2}x$.

   Common mistake: Forgetting the inner derivative ${\sec }^{2}x$.

4. Find the maximum of $f(x)=\sin x+\cos x$ on $[0,2\pi ]$ using derivatives.

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   $f^{\prime }=\cos x-\sin x=0$ at $x=\frac{\pi }{4}$ and $\frac{5\pi }{4}$; $f\left(\frac{\pi }{4}\right)=\sqrt{2}$ is the maximum.

   Common mistake: Choosing $\frac{5\pi }{4}$ (a minimum).

5. Differentiate $y=\arccos \frac{x}{2}$.

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   $-\frac{\frac{1}{2}}{\sqrt{1-\frac{x^2}{4}}}=-\frac{1}{\sqrt{4-x^2}}$.

   Common mistake: Forgetting the minus sign.

6. Compute $\tan \left(\arccos \frac{3}{5}\right)$.

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   If $\cos \theta =\frac{3}{5}$ with $\theta \in [0,\pi ]$, then $\sin \theta =\frac{4}{5}$, so $\tan \theta =\frac{4}{3}$.

   Common mistake: Taking $\sin \theta$ negative.

7. Show that $\frac{d}{dx}(\arcsin x+\arccos x)=0$ and deduce the value of $\arcsin x+\arccos x$.

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   The derivatives cancel, so the sum is constant; at $x=0$ it is $0+\frac{\pi }{2}$. Hence $\arcsin x+\arccos x=\frac{\pi }{2}$.

   Common mistake: Concluding the sum is 0.

8. Find the equation of the tangent to $y=\sin x$ at $x=\pi$.

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   $y(\pi )=0$, $y^{\prime }(\pi )=\cos \pi =-1$: $y=-(x-\pi )=\pi -x$.

   Common mistake: Using $\sin \pi$ as the gradient.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find $\cos \left(\arcsin \frac{3}{5}\right)$.	a. $\pi$
   2. Find the period of $y=3\sin (2x-\pi )$.	b. $[0,6]$
   3. Find the range of $y=2-3\cos x$.	c. $[-1,5]$
   4. Find the domain of $f(x)=\arccos \left(\frac{x}{3}-1\right)$.	d. $\frac{4}{5}$

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   1 → d ($\frac{4}{5}$) · 2 → a ($\pi$) · 3 → c ($[-1,5]$) · 4 → b ($[0,6]$)

   • Step 1. Let $\theta =\arcsin \frac{3}{5}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, so $\cos \theta \ge 0$.
     Step 2. $\cos \theta =\sqrt{1-\frac{9}{25}}=\frac{4}{5}$.
   • Step 1. The period of $\sin (kx+c)$ is $\frac{2\pi }{|k|}$.
     Step 2. $k=2$: period $\pi$.
   • Step 1. $-1\le \cos x\le 1$, so $-3\le -3\cos x\le 3$.
     Step 2. Add 2: $-1\le y\le 5$.
   • Step 1. Need $-1\le \frac{x}{3}-1\le 1$.
     Step 2. $0\le \frac{x}{3}\le 2$, so $0\le x\le 6$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $y=\pi -x$  ·  $0$  ·  $\frac{2}{1+4x^2}$  ·  $\frac{1}{2}$
   1. Find ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$. →
   2. Find the tangent to $y=\sin x$ at $x=\pi$. →
   3. Differentiate $y=\arctan (2x)$. →
   Show answer

   1: $\frac{1}{2}$ · 2: $y=\pi -x$ · 3: $\frac{2}{1+4x^2}$

   • Step 1. $1-\cos x=2{\sin }^2\frac{x}{2}$.
     Step 2. $\frac{2{\sin }^2(x/2)}{x^2}=\frac{1}{2}{\left(\frac{\sin (x/2)}{x/2}\right)}^2\to \frac{1}{2}$.
   • Step 1. Point $(\pi ,0)$; gradient $\cos \pi =-1$.
     Step 2. $y-0=-(x-\pi )$, so $y=\pi -x$.
   • Step 1. $(\arctan u{)}^{\prime }=\frac{u^{\prime }}{1+u^2}$, $u=2x$.
     Step 2. $\frac{2}{1+4x^2}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find $\sin \left(2\arctan \frac{1}{2}\right)$.	a. $\frac{4}{5}$
   2. Find ${\lim }_{x\to 0}\frac{\sin 5x}{2x}$.	b. $-4$
   3. $y=\sin 2x$. Find $y^{\prime \prime }\left(\frac{\pi }{4}\right)$.	c. $\frac{5}{2}$
   4. Find the maximum value of $f(x)=\sin x+\cos x$.	d. $\sqrt{2}$

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   1 → a ($\frac{4}{5}$) · 2 → c ($\frac{5}{2}$) · 3 → b ($-4$) · 4 → d ($\sqrt{2}$)

   • Step 1. Let $\tan \theta =\frac{1}{2}$: $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$.
     Step 2. $\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$.
   • Step 1. $\frac{\sin 5x}{2x}=\frac{5}{2}\times \frac{\sin 5x}{5x}$.
     Step 2. $\frac{\sin 5x}{5x}\to 1$, so the limit is $\frac{5}{2}$.
   • Step 1. $y^{\prime }=2\cos 2x$, $y^{\prime \prime }=-4\sin 2x$.
     Step 2. $-4\sin \frac{\pi }{2}=-4$.
   • Step 1. $f^{\prime }(x)=\cos x-\sin x=0$ at $x=\frac{\pi }{4}$.
     Step 2. $f\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $2\sqrt{3}$  ·  $2$  ·  $8$  ·  $-\frac{\pi }{4}$
   1. Find ${\lim }_{x\to 0}\frac{1-\cos 4x}{x^2}$. →
   2. Find $\arctan (-1)$. →
   3. Find the derivative of $\sec x$ at $x=\frac{\pi }{3}$. →
   Show answer

   1: $8$ · 2: $-\frac{\pi }{4}$ · 3: $2\sqrt{3}$

   • Step 1. $1-\cos 4x=2{\sin }^{2}2x$.
     Step 2. $\frac{2{\sin }^{2}2x}{x^2}=8{\left(\frac{\sin 2x}{2x}\right)}^2\to 8$.
   • Step 1. $\tan \left(-\frac{\pi }{4}\right)=-1$.
     Step 2. The range of $\arctan$ is $\left]-\frac{\pi }{2},\frac{\pi }{2}\right[$, so the answer is $-\frac{\pi }{4}$.
   • Step 1. $(\sec x{)}^{\prime }=\sec x\tan x$.
     Step 2. $\sec \frac{\pi }{3}=2$ and $\tan \frac{\pi }{3}=\sqrt{3}$: $2\sqrt{3}$.

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