Inverse Trigonometric Functions ISC Class 12: Notes, Formulas and Solved Questions

What Inverse Trigonometric Functions is about

Weightage: Inverse Trigonometric Functions is part of the unit Relations and Functions, which carries 10 of the 80 marks in the ISC Mathematics, shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Principal value branches: domain and range

Each inverse function is defined on a restricted branch. ${\sin }^{-1}x$: domain [−1, 1], range $[-\frac{\pi }{2},\frac{\pi }{2}]$. ${\cos }^{-1}x$: domain [−1, 1], range [0, π]. ${\tan }^{-1}x$: domain R, range $(-\frac{\pi }{2},\frac{\pi }{2})$. ${\cot }^{-1}x$: domain R, range (0, π). ${\sec }^{-1}x$: domain |x| ≥ 1, range $[0,\pi ]-\{\frac{\pi }{2}\}$. ${\text{cosec}}^{-1}x$: domain |x| ≥ 1, range $[-\frac{\pi }{2},\frac{\pi }{2}]-\{0\}$. The principal value of an expression is the unique angle in the range whose trigonometric ratio equals the given number. Note that ${\sin }^{-1}x$ does not mean $\frac{1}{\sin x}$.

Worked example

Find the principal values of ${\sin }^{-1}(-\frac{1}{2})$, ${\cos }^{-1}(-\frac{1}{2})$ and ${\tan }^{-1}(-\sqrt{3})$.

Try it first, then show the solutionSolution

1. ${\sin }^{-1}(-\frac{1}{2})$: angle in $[-\frac{\pi }{2},\frac{\pi }{2}]$ with sine −1/2 is $-\frac{\pi }{6}$.
2. ${\cos }^{-1}(-\frac{1}{2})$: angle in [0, π] with cosine −1/2 is $\frac{2\pi }{3}$.
3. ${\tan }^{-1}(-\sqrt{3})=-\frac{\pi }{3}$.

Answer: $-\frac{\pi }{6}$, $\frac{2\pi }{3}$, $-\frac{\pi }{3}$

Graphs of inverse trigonometric functions

The graph of an inverse function is the reflection of its restricted parent in y = x. $y={\sin }^{-1}x$ rises from $(-1,-\frac{\pi }{2})$ through the origin to $(1,\frac{\pi }{2})$. $y={\cos }^{-1}x$ falls from (−1, π) through $(0,\frac{\pi }{2})$ to (1, 0). $y={\tan }^{-1}x$ increases over all of R, through the origin, approaching the horizontal asymptotes $y=\pm \frac{\pi }{2}$. $y={\cot }^{-1}x$ decreases from near π to near 0. The graphs show the domains and ranges at a glance and help in assertion-reason items.

Worked example

From the graph of $y={\sin }^{-1}x$, state its domain and range and whether it is increasing or decreasing. Does ${\sin }^{-1}(\sin x)=x$ for every real x?

Try it first, then show the solutionSolution

1. Domain [−1, 1], range $[-\frac{\pi }{2},\frac{\pi }{2}]$; the curve rises throughout, so it is increasing.
2. ${\sin }^{-1}(\sin x)$ always lies in $[-\frac{\pi }{2},\frac{\pi }{2}]$, so it equals x only for x in that interval.

Answer: Domain [−1, 1], range $[-\pi /2,\pi /2]$, increasing; ${\sin }^{-1}(\sin x)=x$ only for $-\pi /2\le x\le \pi /2$

Elementary properties

Negative arguments: ${\sin }^{-1}(-x)=-{\sin }^{-1}x$, ${\tan }^{-1}(-x)=-{\tan }^{-1}x$, ${\text{cosec}}^{-1}(-x)=-{\text{cosec}}^{-1}x$; but ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$, ${\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$, ${\sec }^{-1}(-x)=\pi -{\sec }^{-1}x$. Reciprocals: ${\sin }^{-1}x={\text{cosec}}^{-1}\frac{1}{x}$, ${\cos }^{-1}x={\sec }^{-1}\frac{1}{x}$, ${\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$ (x > 0). Complementary sums: ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ (|x| ≤ 1), ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$, ${\sec }^{-1}x+{\text{cosec}}^{-1}x=\frac{\pi }{2}$ (|x| ≥ 1). Self-cancelling: $\sin ({\sin }^{-1}x)=x$ for |x| ≤ 1, but ${\sin }^{-1}(\sin \theta )=\theta$ only for θ in the principal range.

Worked example

Find ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$ and ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$.

Try it first, then show the solutionSolution

1. $\frac{2\pi }{3}$ is outside $[-\frac{\pi }{2},\frac{\pi }{2}]$; $\sin \frac{2\pi }{3}=\sin \frac{\pi }{3}$, so the answer is $\frac{\pi }{3}$.
2. ${\sec }^{-1}(-2)=\pi -{\sec }^{-1}2=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$; so $\frac{\pi }{3}-\frac{2\pi }{3}=-\frac{\pi }{3}$.

Answer: $\frac{\pi }{3}$; $-\frac{\pi }{3}$

Converting between inverse functions

Draw a right triangle. If $\theta ={\sin }^{-1}x$ (0 ≤ x < 1), the opposite side is x, hypotenuse 1, adjacent $\sqrt{1-x^2}$. So ${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$. Similarly ${\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}$ for x ≥ 0. This lets you evaluate expressions such as $\cos ({\tan }^{-1}\frac{3}{4})$ quickly and reduce sums to a single function before using a sum formula.

Worked example

Evaluate $\cos \left({\sin }^{-1}\frac{3}{5}\right)$ and $\tan \left({\cos }^{-1}\frac{5}{13}\right)$. Show that ${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$.

Try it first, then show the solutionSolution

1. ${\sin }^{-1}\frac{3}{5}$: triangle 3, 4, 5; cosine = $\frac{4}{5}$ (positive, angle in the first quadrant).
2. ${\cos }^{-1}\frac{5}{13}$: triangle 5, 12, 13; tangent = $\frac{12}{5}$.
3. In the 3-4-5 triangle, tan = 3/4, so ${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$.

Answer: $\frac{4}{5}$; $\frac{12}{5}$; shown

Sum and difference formulas

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$ (xy < 1); ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\frac{x-y}{1+xy}$ (xy > −1). ${\sin }^{-1}x\pm {\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}\pm y\sqrt{1-x^2}\right)$; ${\cos }^{-1}x\pm {\cos }^{-1}y={\cos }^{-1}\left(xy\mp \sqrt{1-x^2}\sqrt{1-y^2}\right)$ (with the usual conditions keeping the sums in the principal range). Strategy for proofs: convert every term to the same function (usually tan⁻¹), combine two at a time, and simplify fractions carefully.

Worked example

Show that ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}$ and that ${\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{8}{17}={\sin }^{-1}\frac{77}{85}$.

Try it first, then show the solutionSolution

1. $xy=\frac{1}{6}<1$: ${\tan }^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}={\tan }^{-1}\frac{5/6}{5/6}={\tan }^{-1}1=\frac{\pi }{4}$.
2. ${\sin }^{-1}\left(\frac{3}{5}\cdot \frac{15}{17}+\frac{8}{17}\cdot \frac{4}{5}\right)={\sin }^{-1}\frac{45+32}{85}={\sin }^{-1}\frac{77}{85}$.

Answer: Both identities hold

Formulas for 2tan⁻¹x, 3tan⁻¹x, 2sin⁻¹x and 2cos⁻¹x

With x = tan θ: $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$ (|x| < 1) $={\sin }^{-1}\frac{2x}{1+x^2}$ (|x| ≤ 1) $={\cos }^{-1}\frac{1-x^2}{1+x^2}$ (x ≥ 0). $3{\tan }^{-1}x={\tan }^{-1}\frac{3x-x^3}{1-3x^2}$ ($|x|<\frac{1}{\sqrt{3}}$). With x = sin θ or cos θ: $2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ and $2{\cos }^{-1}x={\cos }^{-1}(2x^2-1)$ for suitable x. These come straight from the double- and triple-angle formulas of trigonometry, so the substitution is the key step in simplification questions.

Worked example

Prove that $2{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{7}=\frac{\pi }{4}$.

Try it first, then show the solutionSolution

1. $2{\tan }^{-1}\frac{1}{3}={\tan }^{-1}\frac{2/3}{1-1/9}={\tan }^{-1}\frac{2/3}{8/9}={\tan }^{-1}\frac{3}{4}$.
2. ${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{28}}={\tan }^{-1}\frac{25/28}{25/28}={\tan }^{-1}1=\frac{\pi }{4}$.

Answer: Proved

Solving equations

Combine the inverse terms into one using the formulas, take the trigonometric ratio of both sides, and solve the resulting algebraic equation. Then check every root in the original equation: a root may violate the condition of the formula used (such as xy < 1) or put a principal value outside its range. Useful shortcuts: if an equation contains both ${\sin }^{-1}x$ and ${\cos }^{-1}x$, replace one using ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$.

Worked example

Solve ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$.

Try it first, then show the solutionSolution

1. ${\tan }^{-1}\frac{5x}{1-6x^2}=\frac{\pi }{4}\Rightarrow \frac{5x}{1-6x^2}=1\Rightarrow 6x^2+5x-1=0$
2. $(6x-1)(x+1)=0$: $x=\frac{1}{6}$ or $x=-1$.
3. x = −1 makes both terms negative, so the left side is negative, not π/4 (and xy = 6 > 1): reject. x = 1/6 works.

Answer: $x=\frac{1}{6}$

Maths in real life: Inverse Trigonometric Functions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Inverse Trigonometric Identities
• Solving Inverse Trigonometric Equations
• Principal Values of Inverse Trigonometric Functions

Inverse Trigonometric Functions formula sheet

Principal ranges

${\sin }^{-1}:[-\frac{\pi }{2},\frac{\pi }{2}];{\cos }^{-1}:[0,\pi ];{\tan }^{-1}:(-\frac{\pi }{2},\frac{\pi }{2});{\cot }^{-1}:(0,\pi )$

sec⁻¹: [0, π] − {π/2}; cosec⁻¹: [−π/2, π/2] − {0}.

Negative arguments

${\sin }^{-1}(-x)=-{\sin }^{-1}x,{\cos }^{-1}(-x)=\pi -{\cos }^{-1}x,{\tan }^{-1}(-x)=-{\tan }^{-1}x$

cot⁻¹ and sec⁻¹ follow the cos⁻¹ pattern.

Complementary sums

${\sin }^{-1}x+{\cos }^{-1}x={\tan }^{-1}x+{\cot }^{-1}x={\sec }^{-1}x+{\text{cosec}}^{-1}x=\frac{\pi }{2}$

Within each domain.

Reciprocals

${\sin }^{-1}x={\text{cosec}}^{-1}\frac{1}{x},{\cos }^{-1}x={\sec }^{-1}\frac{1}{x},{\tan }^{-1}x={\cot }^{-1}\frac{1}{x}(x>0)$

Conversion

${\sin }^{-1}x={\cos }^{-1}\sqrt{1-x^2}={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$

For 0 ≤ x < 1.

Sum and difference of tan⁻¹

${\tan }^{-1}x\pm {\tan }^{-1}y={\tan }^{-1}\frac{x\pm y}{1\mp xy}$

Sum: xy < 1; difference: xy > −1.

Sum and difference of sin⁻¹ and cos⁻¹

${\sin }^{-1}x\pm {\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}\pm y\sqrt{1-x^2});{\cos }^{-1}x\pm {\cos }^{-1}y={\cos }^{-1}(xy\mp \sqrt{1-x^2}\sqrt{1-y^2})$

Check that the result stays in the principal range.

Double angle in tan⁻¹

$2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}={\sin }^{-1}\frac{2x}{1+x^2}={\cos }^{-1}\frac{1-x^2}{1+x^2}$

|x| < 1; |x| ≤ 1; x ≥ 0 respectively.

Triple angle

$3{\tan }^{-1}x={\tan }^{-1}\frac{3x-x^3}{1-3x^2}$

$|x|<1/\sqrt{3}$.

Double angle in sin⁻¹ and cos⁻¹

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

For suitable x (e.g. |x| ≤ 1/√2 for the first).

How the board asks Inverse Trigonometric Functions

The question patterns to expect, and the method that works for each.

Tricks to remember Inverse Trigonometric Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Odd ones keep the minus

  sin⁻¹, tan⁻¹, cosec⁻¹ pull the minus out; cos⁻¹, cot⁻¹, sec⁻¹ give π minus.

  Use it when: Negative-argument values.

• Complementary pairs add to π/2

  sin-cos, tan-cot, sec-cosec: each pair sums to a right angle.

  Use it when: Replacing one inverse by the other in equations.

• tan⁻¹ sum is tan(A + B)

  $\frac{x+y}{1-xy}$ is just tan(A + B) with tan A = x, tan B = y.

  Use it when: Remembering the sum formula and its condition.

• Triangle trick

  Draw a right triangle with the given ratio to convert sin⁻¹ ↔ cos⁻¹ ↔ tan⁻¹.

  Use it when: Evaluations like cos(tan⁻¹ x).

• Check the roots

  Solved an inverse-trig equation? Substitute back: some roots are imposters.

  Use it when: Equation questions.

Inverse Trigonometric Functions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Inverse Trigonometric Functions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Not knowing domain and range of sin⁻¹

  CISCE examiners' comments, 2025: in a graph-based question, many candidates could not write the domain and range of $y={\sin }^{-1}x$.

  Wrong working

  Domain R, range [0, π].

  Right working

  Domain [−1, 1], range $[-\frac{\pi }{2},\frac{\pi }{2}]$.

  ✓Do this: Learn the six domain-range pairs as one table, with a sketch of each graph.

  Usually costs: 1 mark.

• ✗Mistake: Missing the basic complementary property

  CISCE examiners' comments, 2025: given $({\tan }^{-1}x{)}^2-({\cot }^{-1}x{)}^2$, candidates failed to use ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$.

  Wrong working

  Expands each square separately and gets stuck.

  Right working

  Factorise as $({\tan }^{-1}x+{\cot }^{-1}x)({\tan }^{-1}x-{\cot }^{-1}x)=\frac{\pi }{2}({\tan }^{-1}x-{\cot }^{-1}x)$.

  ✓Do this: Look for sin-cos, tan-cot, sec-cosec pairs first.

  Usually costs: 1-2 marks.

• ✗Mistake: Choosing a long route and losing the algebra

  CISCE examiners' comments, 2024: in solving inverse-trigonometric equations, candidates knew the identities but made simplification errors or used lengthy calculations and could not finish.

  Wrong working

  Converts every term to sin⁻¹ with nested square roots.

  Right working

  Use the complementary sum or convert to tan⁻¹, which keeps the algebra rational.

  ✓Do this: Spend ten seconds choosing the identity that makes the algebra simplest.

  Usually costs: 2-3 marks.

• ✗Mistake: Combining terms in the wrong order

  CISCE examiners' comments, 2024: in proving a relation from ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\pi$, candidates took an unhelpful combination of terms and could not reach the result.

  Wrong working

  Combines ${\sin }^{-1}x+{\sin }^{-1}y$ when the target isolates x.

  Right working

  Write ${\sin }^{-1}y+{\sin }^{-1}z=\pi -{\sin }^{-1}x$, then take sine or cosine of both sides.

  ✓Do this: Isolate the term that stands alone in the required result.

  Usually costs: 2-3 marks.

• ✗Mistake: Keeping a root that fails the formula's condition

  A frequent error in tan⁻¹ equations: both roots of the quadratic are given without checking.

  Wrong working

  x = 1/6 or x = −1.

  Right working

  x = −1 gives a negative left side, so it is rejected; x = 1/6 only.

  ✓Do this: Substitute every root back before writing the answer.

  Usually costs: 1 mark.

Practice questions on Inverse Trigonometric Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The range of the principal value branch of ${\cos }^{-1}x$ is:
   (a) $[-\frac{\pi }{2},\frac{\pi }{2}]$
   (b) [0, π]
   (c) (0, π)
   (d) $(-\frac{\pi }{2},\frac{\pi }{2})$

   Hint

   cos is one-one on this interval.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The value of ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)$ is:
   (a) π
   (b) $-\frac{\pi }{3}$
   (c) $\frac{\pi }{3}$
   (d) $\frac{2\pi }{3}$

   Hint

   ${\sec }^{-1}(-x)=\pi -{\sec }^{-1}x$.

   Answer

   (b)

3. Q3 · 1 mark · MCQBoard level

   The domain of ${\sin }^{-1}(2x-1)$ is:
   (a) [−1, 1]
   (b) [0, 1]
   (c) [−1, 0]
   (d) [0, 2]

   Hint

   −1 ≤ 2x − 1 ≤ 1.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Write the principal value of ${\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

   Hint

   Odd function.

   Answer

   $-\frac{\pi }{3}$

5. Q5 · 2 marks · Short answerBoard level

   Evaluate $\cos \left({\sin }^{-1}\frac{3}{5}\right)$ and $\tan \left({\cos }^{-1}\frac{5}{13}\right)$.

   Hint

   Right triangles 3-4-5 and 5-12-13.

   Answer

   4/5 and 12/5

6. Q6 · 2 marks · Short answerHarder board level

   Show that ${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$.

   Hint

   Same triangle.

   Answer

   If θ = ${\sin }^{-1}\frac{3}{5}$ (acute), cos θ = 4/5, so tan θ = 3/4; hence θ = ${\tan }^{-1}\frac{3}{4}$.

7. Q7 · 2 marks · Short answerHarder board level

   Prove that ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}$.

   Hint

   Product 1/6 < 1.

   Answer

   ${\tan }^{-1}\frac{5/6}{5/6}={\tan }^{-1}1=\frac{\pi }{4}$.

8. Q8 · 2 marks · Short answerHOTS / competency

   Evaluate $\tan \left(2{\tan }^{-1}\frac{1}{5}-\frac{\pi }{4}\right)$.

   Hint

   Double first, then subtract π/4.

   Answer

   −7/17

9. Q9 · 3 marks · Short answerHarder board level

   Write ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ (x ≠ 0) in its simplest form.

   Hint

   x = tan θ.

   Answer

   $\frac{1}{2}{\tan }^{-1}x$

10. Q10 · 3 marks · Short answerHarder board level

    Prove that ${\tan }^{-1}x+{\tan }^{-1}\frac{2x}{1-x^2}={\tan }^{-1}\frac{3x-x^3}{1-3x^2}$ for $|x|<\frac{1}{\sqrt{3}}$.

    Hint

    The second term is 2 tan⁻¹ x.

    Answer

    With x = tan θ, the left side is θ + 2θ = 3θ, and $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$; the condition keeps 3θ in $(-\frac{\pi }{2},\frac{\pi }{2})$.

11. Q11 · 3 marks · Short answerHarder board level

    Solve $2{\tan }^{-1}(\cos x)={\tan }^{-1}(2\text{cosec}x)$, $0<x<\pi$.

    Hint

    Use the 2tan⁻¹ formula on the left.

    Answer

    $\frac{2\cos x}{1-{\cos }^{2}x}=\frac{2}{\sin x}\Rightarrow \frac{\cos x}{{\sin }^{2}x}=\frac{1}{\sin x}\Rightarrow \cos x=\sin x\Rightarrow x=\frac{\pi }{4}$.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Sketch $y={\cos }^{-1}x$ and state its domain and range. [2]
    (b) Prove that ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$. [3]

    Hint

    Find the sines first.

    Answer

    (a) Decreasing curve from (−1, π) through (0, π/2) to (1, 0); domain [−1, 1], range [0, π]. (b) sin A = 3/5, sin B = 5/13; cos(A + B) = 48/65 − 15/65 = 33/65, and A + B ∈ [0, π], so A + B = ${\cos }^{-1}\frac{33}{65}$.

13. Q13 · 2 marks · Short answerHarder board level

    If ${\sin }^{-1}x=\frac{\pi }{5}$ for some x ∈ [−1, 1], find ${\cos }^{-1}x$.

    Hint

    Complementary sum.

    Answer

    ${\cos }^{-1}x=\frac{\pi }{2}-\frac{\pi }{5}=\frac{3\pi }{10}$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Range of ${\tan }^{-1}x$?Show answer

   $(-\frac{\pi }{2},\frac{\pi }{2})$.

2. 2.${\cos }^{-1}(-x)$ = ?Show answer

   $\pi -{\cos }^{-1}x$.

3. 3.${\sin }^{-1}x+{\cos }^{-1}x$ = ?Show answer

   $\frac{\pi }{2}$ for |x| ≤ 1.

4. 4.${\tan }^{-1}x+{\tan }^{-1}y$ = ? (xy < 1)Show answer

   ${\tan }^{-1}\frac{x+y}{1-xy}$.

5. 5.$2{\tan }^{-1}x$ as a sin⁻¹?Show answer

   ${\sin }^{-1}\frac{2x}{1+x^2}$ for |x| ≤ 1.

6. 6.$3{\tan }^{-1}x$ = ?Show answer

   ${\tan }^{-1}\frac{3x-x^3}{1-3x^2}$.

7. 7.Is ${\sin }^{-1}x$ the same as $(\sin x{)}^{-1}$?Show answer

   No; ${\sin }^{-1}x$ is the inverse function, $(\sin x{)}^{-1}=\text{cosec}x$.

8. 8.When does ${\sin }^{-1}(\sin \theta )=\theta$?Show answer

   Only for θ in $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state the domain and range of all six inverse functions?
• ☐Can you sketch ${\sin }^{-1}x$, ${\cos }^{-1}x$ and ${\tan }^{-1}x$?
• ☐Can you find principal values with negative arguments?
• ☐Can you use the complementary sums to simplify?
• ☐Can you convert sin⁻¹ to tan⁻¹ with a triangle?
• ☐Can you prove a three-term tan⁻¹ identity?
• ☐Can you simplify using x = tan θ?
• ☐Can you solve an inverse-trig equation and reject false roots?

Inverse Trigonometric Functions: frequently asked questions