Tutorial 6: Inverse Trigonometric Functions

Tutorial 6: Inverse Trigonometric Functions

1. Domain of Inverse Trigonometric Functions

Consider two inverse trigonometric functions:

• $f_1(x) = \tan^{-1} x$
• $f_2(x) = \cos^{-1} 3x$

Domain of $f_1(x) = \tan^{-1} x$:

• All real numbers: $(-\infty, \infty)$

Domain of $f_2(x) = \cos^{-1} 3x$:

• The domain of $\cos^{-1} u$ is $-1 \leq u \leq 1$.
• So, $-1 \leq 3x \leq 1$ gives $-\frac{1}{3} \leq x \leq \frac{1}{3}$.

Domain of $g(x) = f_1(x) - f_2(x)$:

• Intersection of the two domains: $[-\frac{1}{3}, \frac{1}{3}]$

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2. Evaluating Trigonometric Functions from Inverse Tangent

Given $\theta = \tan^{-1} \left( \frac{4}{3} \right)$:

• Construct a right triangle with opposite = 4, adjacent = 3.
• Hypotenuse = $\sqrt{3^2 + 4^2} = 5$.

Exact values:

• $\sin \theta = \frac{4}{5}$
• $\cos \theta = \frac{3}{5}$
• $\cot \theta = \frac{3}{4}$
• $\sec \theta = \frac{5}{3}$
• $\csc \theta = \frac{5}{4}$

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3. Evaluating Trigonometric Functions from Inverse Secant

Given $\theta = \sec^{-1}(2.6) = \sec^{-1} \left( \frac{13}{5} \right)$:

• $\sec \theta = \frac{13}{5}$, so $\cos \theta = \frac{5}{13}$.
• Construct a right triangle with adjacent = 5, hypotenuse = 13.
• Opposite = $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.

Exact values:

• $\sin \theta = \frac{12}{13}$
• $\cos \theta = \frac{5}{13}$
• $\cot \theta = \frac{5}{12}$
• $\sec \theta = \frac{13}{5}$
• $\csc \theta = \frac{13}{12}$

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4. Evaluating Exact Values of Composite Expressions

a) $\sin \left( 2 \tan^{-1} \frac{2}{3} \right)$

• Let $\theta = \tan^{-1} \frac{2}{3}$, so $\tan \theta = \frac{2}{3}$.
• Use identity: $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \cdot \frac{2}{3}}{1 + \left( \frac{2}{3} \right)^2} = \frac{\frac{4}{3}}{1 + \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{13}{9}} = \frac{4}{3} \cdot \frac{9}{13} = \frac{12}{13}$

b) $\cos \left( \sin^{-1} \frac{3}{5} + \sec^{-1} 2 \right)$

• Let $\theta = \sin^{-1} \frac{3}{5}$, so $\sin \theta = \frac{3}{5}$, $\cos \theta = \frac{4}{5}$.
• Let $\phi = \sec^{-1} 2$, so $\sec \phi = 2$, $\cos \phi = \frac{1}{2}$, $\sin \phi = \frac{\sqrt{3}}{2}$.
• Use identity: $\cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi = \frac{4}{5} \cdot \frac{1}{2} - \frac{3}{5} \cdot \frac{\sqrt{3}}{2} = \frac{2}{5} - \frac{3\sqrt{3}}{10} = \frac{4 - 3\sqrt{3}}{10}$

c) $\tan \left( \frac{\pi}{4} + \cot^{-1} \frac{4}{5} \right)$

• Let $\theta = \cot^{-1} \frac{4}{5}$, so $\cot \theta = \frac{4}{5}$, $\tan \theta = \frac{5}{4}$.
• Use identity: $\tan \left( \frac{\pi}{4} + \theta \right) = \frac{1 + \tan \theta}{1 - \tan \theta} = \frac{1 + \frac{5}{4}}{1 - \frac{5}{4}} = \frac{\frac{9}{4}}{-\frac{1}{4}} = -9$

d) $\sin \left( \tan^{-1} \sqrt{3} + \cos^{-1} \frac{1}{\sqrt{2}} \right)$

• $\tan^{-1} \sqrt{3} = \frac{\pi}{3}$, $\cos^{-1} \frac{1}{\sqrt{2}} = \frac{\pi}{4}$.
• Compute: $\sin \left( \frac{\pi}{3} + \frac{\pi}{4} \right) = \sin \frac{7\pi}{12} = \sin 105^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$

e) $\sin \left( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \right)$

• Let $\theta = \sec^{-1} \frac{2}{\sqrt{3}}$, so $\sec \theta = \frac{2}{\sqrt{3}}$, $\cos \theta = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{6}$.
• $\sin \theta = \sin \frac{\pi}{6} = \frac{1}{2}$

f) $\cos \left( 2 \sin^{-1} \frac{\sqrt{3}}{5} + \frac{\pi}{2} \right)$

• Let $\theta = \sin^{-1} \frac{\sqrt{3}}{5}$, so $\sin \theta = \frac{\sqrt{3}}{5}$, $\cos \theta = \frac{\sqrt{22}}{5}$.
• Use $\cos(2\theta + \frac{\pi}{2}) = -\sin 2\theta$.
• $\sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{3}}{5} \cdot \frac{\sqrt{22}}{5} = \frac{2\sqrt{66}}{25}$.
• Result: $-\frac{2\sqrt{66}}{25}$

g) $\cos \left( \tan^{-1} \frac{4}{3} - \csc^{-1} \frac{3}{\sqrt{5}} \right)$

• Let $\theta = \tan^{-1} \frac{4}{3}$, so $\tan \theta = \frac{4}{3}$, $\sin \theta = \frac{4}{5}$, $\cos \theta = \frac{3}{5}$.
• Let $\phi = \csc^{-1} \frac{3}{\sqrt{5}}$, so $\csc \phi = \frac{3}{\sqrt{5}}$, $\sin \phi = \frac{\sqrt{5}}{3}$, $\cos \phi = \frac{2}{3}$.
• Use identity: $\cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi = \frac{3}{5} \cdot \frac{2}{3} + \frac{4}{5} \cdot \frac{\sqrt{5}}{3} = \frac{2}{5} + \frac{4\sqrt{5}}{15} = \frac{6 + 4\sqrt{5}}{15}$

h) $\csc(\sec^{-1} 3) + \tan \left( \cos^{-1} \frac{1}{2} \right)$

• For $\sec^{-1} 3$: $\sec \theta = 3$, $\cos \theta = \frac{1}{3}$, $\sin \theta = \frac{2\sqrt{2}}{3}$, $\csc \theta = \frac{3}{2\sqrt{2}}$.
• For $\cos^{-1} \frac{1}{2}$: $\cos \phi = \frac{1}{2}$, $\phi = \frac{\pi}{3}$, $\tan \phi = \sqrt{3}$.
• Sum: $\frac{3}{2\sqrt{2}} + \sqrt{3}$

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5. Simplifying Trigonometric Expressions with Intervals

a) $\cos \left( \sin^{-1}

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Source: FAC1004 Tutorial 6 25-26.pdf