Simplify Inverse Trigonometric Functions

Learn how to simplify expressions involving inverse trigonometric functions in Grade 12 math. This page includes practice questions with step-by-step solutions and clear explanations to help you master this important topic.

Questions with Solutions

Question 1

Simplify the expressions:
  1. )\( \sin(\arcsin(x)) \quad \text{and} \quad \arcsin(\sin(x)) \)
  2. )\( \cos(\arccos(x)) \quad \text{and} \quad \arccos(\cos(x)) \)
  3. )\( \tan(\arctan(x)) \quad \text{and} \quad \arctan(\tan(x)) \)

Solution

  1. )

    sin and arcsin are inverse of each other and therefore the properties of inverse functions may be used to write : \[ \sin(\arcsin(x)) = x, \quad \text{for } -1 \leq x \leq 1 \] \[ \arcsin(\sin(x)) = x, \quad \text{for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

    NOTE:

    If \( x \) in \( \arcsin(\sin(x)) \) is not in the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), find \( \theta \) in the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) such that \[ \sin(x) = \sin(\theta) \] and then simplify: \[ \arcsin(\sin(x)) = \theta \]

  2. )

    Cosine and arccosine are inverse functions of each other. Therefore, the properties of inverse functions can be used to write: \[ \cos(\arccos(x)) = x, \quad \text{for } -1 \leq x \leq 1 \] \[ \arccos(\cos(x)) = x, \quad \text{for } x \in [0, \pi] \]

    NOTE:

    If \( x \) in \( \arccos(\cos(x)) \) is not in the interval \([0, \pi]\), find \( \theta \) in the interval \([0, \pi] \) such that \( \cos(x) = \cos(\theta) \), and then simplify \( \arccos(\cos(x)) = \theta \).

  3. )

    tan and arctan are inverse of each other and therefore the properties of inverse functions may be used to write

    \[ \tan(\arctan(x)) = x \] \[ \arctan(\tan(x)) = x \quad \text{for } x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \]

    NOTE:

    If \( x \) in \( \arctan(\tan(x)) \) is not in the interval \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), find \( \theta \) in the interval \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \) such that \( \tan(x) = \tan(\theta) \) and then simplify \( \arctan(\tan(x)) = \theta \)

Question 2

Express the following as algebraic expressions: \[ \sin(\arccos(x)) \] and \[ \tan(\arccos(x)) \]

Solution

Let \( A = \arccos(x) \). Hence, \[ \cos(A) = \cos(\arccos(x)) = x \]

Use a right triangle with angle \( A \) such that \( \cos(A) = x \) (or \( \frac{x}{1} \)). Find the second leg and calculate \( \sin(A) \) and \( \tan(A) \).

Right triangle for inverse cosine function

\[ \sin(\arccos(x)) = \sin(A) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} \quad \text{for } x \in [-1, 1] \] \[ \tan(\arccos(x)) = \tan(A) = \frac{\sqrt{1 - x^2}}{x} \quad \text{for } x \in [-1, 0) \cup (0, 1] \]

Question 3

Express the following as algebraic expressions: \[ \cos(\arcsin(x)) \] and \[ \tan(\arcsin(x)) \]

Solution

Let \( A = \arcsin(x) \). Hence, \[ \sin(A) = \sin(\arcsin(x)) = x \]

Use a right triangle with angle \( A \) such that \( \sin(A) = x \) (or \( \frac{x}{1} \)), find the second leg and calculate \( \cos(A) \) and \( \tan(A) \). Right triangle illustrating sin(arcsin(x)) = x \[ \cos(\arcsin(x)) = \cos(A) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} \quad \text{for } x \in [-1, 1] \] \[ \tan(\arcsin(x)) = \tan(A) = \frac{x}{\sqrt{1 - x^2}} \quad \text{for } x \in (-1, 1) \]

Question 4

Express the following as algebraic expressions: \[ \sin(\arctan(x)) \] and \[ \cos(\arctan(x)) \]

Solution

Let \( A = \arctan(x) \). Hence, \[ \tan(A) = \tan(\arctan(x)) = x \]

Use a right triangle with angle \( A \) such that \( \tan(A) = x \) (or \( \frac{x}{1} \)). Find the hypotenuse and calculate \( \sin(A) \) and \( \cos(A) \). Right triangle illustrating angle A with tan(A) = x \[ \sin(\arctan(x)) = \sin(A) = \frac{x}{\sqrt{1 + x^2}} \] \[ \cos(\arctan(x)) = \cos(A) = \frac{1}{\sqrt{1 + x^2}} \]

Question 5

Simplify the following expressions:

  1. ) arccos(0) , arcsin(-1) , arctan(-1)
  2. ) sin(arcsin(-1/2)) , arccos(cos(π/2)) , arccos(cos(-π/2))
  3. ) cos(arcsin(-1/2)) , arcsin(sin(π/3)) , arcsin(tan(3π/4)) d) arccos(tan(7π/4)) , arcsin(sin(13π/3)) , arctan(tan(-17π/4)) , arcsin(sin(9π/5))

Solution

  1. )

    Use definition. \[ \arccos(0) = \frac{\pi}{2} \quad \text{because} \quad \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \frac{\pi}{2} \in [0, \pi] \] \[ \arcsin(-1) = -\frac{\pi}{2} \quad \text{because} \quad \sin\left(-\frac{\pi}{2}\right) = -1 \quad \text{and} \quad -\frac{\pi}{2} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] \[ \arctan(-1) = -\frac{\pi}{4} \quad \text{because} \quad \tan\left(-\frac{\pi}{4}\right) = -1 \quad \text{and} \quad -\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]

  2. )

    Simplify inner functions then the outer functions using definitions. \[ \sin(\arcsin(-\frac{1}{2})) = \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \] \[ \arccos(\cos(\frac{\pi}{2})) = \arccos(0) = \frac{\pi}{2} \] \[ \arccos(\cos(-\frac{\pi}{2})) = \arccos(0) = \frac{\pi}{2} \]

  3. )

    Simplify inner functions then the outer functions using definitions. \[ \cos(\arcsin(-\frac{1}{2})) = \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] \[ \arcsin(\sin(\frac{\pi}{3})) = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] \[ \arcsin(\tan(\frac{3\pi}{4})) = \arcsin(-1) = -\frac{\pi}{2} \]

  4. )

    Simplify inner functions then the outer functions using definitions. \[ \arccos\big(\tan(7\pi/4)\big) = \arccos(-1) = \pi \] \[ \arcsin\big(\sin(13\pi/3)\big) = \arcsin\big(\sin(4\pi + \pi/3)\big) = \arcsin\big(\sin(\pi/3)\big) = \pi/3 \] \[ \arctan\big(\tan(-17\pi/4)\big) = \arctan\big(\tan(-4\pi - \pi/4)\big) = \arctan\big(\tan(-\pi/4)\big) = -\pi/4 \] \[ \arcsin\big(\sin(9\pi/5)\big) = \arcsin\big(\sin(2\pi - \pi/5)\big) = \arcsin\big(\sin(-\pi/5)\big) = -\pi/5 \]

Question 6

Let \( A = \arcsin(2/3) \) and \( B = \arccos(-1/2) \). Find the exact value of \( \sin(A + B) \).

Solution

Use the identity \[ \sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) \] to expand the given expression. \[ \sin(A + B) = \sin(\arcsin(\tfrac{2}{3}))\cos(\arccos(-\tfrac{1}{2})) + \cos(\arcsin(\tfrac{2}{3}))\sin(\arccos(-\tfrac{1}{2})) \] Use the above identities to simplify each term in the above expression. \[ \sin(\arcsin(\tfrac{2}{3})) = \tfrac{2}{3} \] (we have used \(\sin(\arcsin(x)) = x\)) \[ \cos(\arccos(-\tfrac{1}{2})) = -\tfrac{1}{2} \] (we have used \(\cos(\arccos(x)) = x\)) \[ \cos(\arcsin(\tfrac{2}{3})) = \sqrt{1 - \left(\tfrac{2}{3}\right)^2} = \sqrt{\tfrac{5}{9}} = \tfrac{\sqrt{5}}{3} \] (we have used \(\cos(\arcsin(x)) = \sqrt{1 - x^2}\)) \[ \sin(\arccos(-\tfrac{1}{2})) = \sqrt{1 - \left(-\tfrac{1}{2}\right)^2} = \sqrt{\tfrac{3}{4}} = \tfrac{\sqrt{3}}{2} \] (we have used \(\sin(\arccos(x)) = \sqrt{1 - x^2}\)) Substitute and calculate. \[ \sin(A + B) = \left(\tfrac{2}{3}\right)\left(-\tfrac{1}{2}\right) + \left(\tfrac{\sqrt{5}}{3}\right)\left(\tfrac{\sqrt{3}}{2}\right) = -\tfrac{1}{3} + \tfrac{\sqrt{15}}{6} \]

Question 7

Write \( Y = \sin(2 \arcsin(x)) \) as an algebraic expression.

Solution

Let \( A = \arcsin(x) \). Hence \( Y \) may be written as \[ Y = \sin(2A) \] Use the identity \[ \sin(2A) = 2 \sin(A) \cos(A) \] to rewrite \( Y \) as follows: \[ Y = 2 \sin(A) \cos(A) = 2 \sin(\arcsin(x)) \cos(\arcsin(x)) \] Use the identities \[ \sin(\arcsin(x)) = x \quad \text{and} \quad \cos(\arcsin(x)) = \sqrt{1 - x^2} \] to rewrite \( Y \) as follows: \[ Y = 2 x \sqrt{1 - x^2} \]

Question 8

Find the exact value of \( Y = \sin(2 \arctan(3/4)) \).

Solution

Let \( A = \arctan\left(\frac{3}{4}\right) \). Hence \( Y \) may be written as \[ Y = \sin(2A) = 2 \sin(A) \cos(A) \] \[ \sin(A) = \sin\left(\arctan\left(\frac{3}{4}\right)\right) = \frac{\frac{3}{4}}{\sqrt{1 + \left(\frac{3}{4}\right)^2}} = \frac{3}{5} \] \[ \cos(A) = \cos\left(\arctan\left(\frac{3}{4}\right)\right) = \frac{1}{\sqrt{1 + \left(\frac{3}{4}\right)^2}} = \frac{4}{5} \] \[ Y = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25} \]

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