Question Statement

Find the unknown angles and use a calculator to evaluate the following as real numbers to three decimal places:

(i) ${\cos }^{-1}\left(\frac{3}{5}\right)={\sin }^{-1}(\dots )$ (ii) ${\sin }^{-1}\left(\frac{2}{3}\right)={\cos }^{-1}(\dots )$ (iii) ${\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right)=-{\cos }^{-1}(\dots )$ (iv) ${\tan }^{-1}\left(\frac{1}{4}\right)={\cos }^{-1}(\dots )$ (v) ${\tan }^{-1}(-1.2)=-{\cos }^{-1}(\dots )$ (vi) ${\cot }^{-1}\left(\frac{-3}{4}\right)=-{\sin }^{-1}(\dots )$ (vii) ${\sec }^{-1}(2.041)={\tan }^{-1}(\dots )$ (viii) ${\sec }^{-1}(-\sqrt{5})={\cot }^{-1}(\dots )$ (ix) ${\csc }^{-1}(1.172)={\sin }^{-1}(\dots )$ (x) ${\csc }^{-1}\left(\frac{-5}{3}\right)={\tan }^{-1}(\dots )$

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Background and Explanation

To solve these problems, we use the relationship between inverse trigonometric functions and right-angled triangles. By setting the given expression equal to an angle $\theta$, we can identify two sides of a triangle (Perpendicular $P$, Base $B$, or Hypotenuse $H$) and use the Pythagorean theorem ($H^2=B^2+P^2$) to find the third side, allowing us to express the angle using a different trigonometric ratio.

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Solution

Part (i)

Let $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$. This implies $\cos \theta =\frac{3}{5}$. In a right-angled triangle, $\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}$, so $B=3$ and $H=5$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =B^2+P^2\\ 5^2& =3^2+P^2\\ 25-9& =P^2\\ P^2& =16⟹P=4\end{array}$ Now, $\sin \theta =\frac{P}{H}=\frac{4}{5}$. Therefore, $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$. Calculator evaluation: $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)\approx 53.130^{\circ }$.

Part (ii)

Let $\theta ={\sin }^{-1}\left(\frac{2}{3}\right)$. This implies $\sin \theta =\frac{2}{3}$. Here, $P=2$ and $H=3$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =B^2+P^2\\ 3^2& =B^2+2^2\\ 9-4& =B^2\\ B^2& =5⟹B=\sqrt{5}\end{array}$ Now, $\cos \theta =\frac{B}{H}=\frac{\sqrt{5}}{3}$. Therefore, ${\sin }^{-1}\left(\frac{2}{3}\right)={\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)$. Calculator evaluation: $\theta ={\sin }^{-1}\left(\frac{2}{3}\right)\approx 41.810^{\circ }$.

Part (iii)

Let $-\theta ={\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right)$. This implies $\sin (-\theta )=\frac{-1}{\sqrt{5}}$, so $\sin \theta =\frac{1}{\sqrt{5}}$. Here, $P=1$ and $H=\sqrt{5}$. Using the Pythagorean theorem: $\begin{array}{rl}{B}^2& =H^2-P^2\\ B^2& =(\sqrt{5}{)}^2-1^2=5-1=4\\ B& =2\end{array}$ Now, $\cos \theta =\frac{B}{H}=\frac{2}{\sqrt{5}}$, so $\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$. Thus, $-\theta =-{\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$. Calculator evaluation: $\theta ={\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right)\approx -26.565^{\circ }$.

Part (iv)

Let $\theta ={\tan }^{-1}\left(\frac{1}{4}\right)$. This implies $\tan \theta =\frac{1}{4}$. Here, $P=1$ and $B=4$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =B^2+P^2\\ H^2& =4^2+1^2=16+1=17\\ H& =\sqrt{17}\end{array}$ Now, $\cos \theta =\frac{B}{H}=\frac{4}{\sqrt{17}}$. Therefore, ${\tan }^{-1}\left(\frac{1}{4}\right)={\cos }^{-1}\left(\frac{4}{\sqrt{17}}\right)$. Calculator evaluation: $\theta ={\tan }^{-1}\left(\frac{1}{4}\right)\approx 14.036^{\circ }$.

Part (v)

Let $-\theta ={\tan }^{-1}(-1.2)$. This implies $\tan \theta =1.2=\frac{1.2}{1}$. Here, $P=1.2$ and $B=1$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =1^2+1.2^2=1+1.44=2.44\\ H& =\sqrt{2.44}=\frac{\sqrt{61}}{5}\end{array}$ Now, $\cos \theta =\frac{B}{H}=\frac{1}{\sqrt{61}/5}=\frac{5}{\sqrt{61}}$. Thus, $-\theta =-{\cos }^{-1}\left(\frac{5}{\sqrt{61}}\right)$. Calculator evaluation: $\theta ={\tan }^{-1}(-1.2)\approx -50.194^{\circ }$.

Part (vi)

Let $-\theta ={\cot }^{-1}\left(\frac{-3}{4}\right)$. This implies $\cot \theta =\frac{3}{4}$. Since $\cot \theta =\frac{B}{P}$, we have $B=3$ and $P=4$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =3^2+4^2=25\\ H& =5\end{array}$ Now, $\sin \theta =\frac{P}{H}=\frac{4}{5}$, so $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$. Thus, $-\theta =-{\sin }^{-1}\left(\frac{4}{5}\right)$. Calculator evaluation: $\theta ={\cot }^{-1}\left(\frac{-3}{4}\right)\approx -53.130^{\circ }$.

Part (vii)

Let $\theta ={\sec }^{-1}(2.041)$. This implies $\sec \theta =2.041=\frac{2041}{1000}$. Since $\sec \theta =\frac{H}{B}$, we have $H=2041$ and $B=1000$. Using the Pythagorean theorem: $\begin{array}{rl}{P}^2& =H^2-B^2\\ P^2& =2041^2-1000^2=3165681\\ P& =\sqrt{3165681}\approx 1779.236\end{array}$ Now, $\tan \theta =\frac{P}{B}=\frac{1779.236}{1000}=1.779236$. Therefore, ${\sec }^{-1}(2.041)={\tan }^{-1}(1.779)$. Calculator evaluation: $\theta ={\sec }^{-1}(2.041)\approx 60.662^{\circ }$.

Part (viii)

Let $\theta ={\sec }^{-1}(-\sqrt{5})$. This implies $\sec \theta =-\sqrt{5}$, so $\cos \theta =\frac{-1}{\sqrt{5}}$. Using the Pythagorean theorem: $\begin{array}{rl}{H}^2& =B^2+P^2\\ (\sqrt{5}{)}^2& =(-1{)}^2+P^2\\ 5-1& =P^2⟹P=2\end{array}$ Now, $\cot \theta =\frac{B}{P}=\frac{-1}{2}$. Therefore, ${\sec }^{-1}(-\sqrt{5})={\cot }^{-1}\left(-\frac{1}{2}\right)$. Calculator evaluation: $\theta ={\sec }^{-1}(-\sqrt{5})\approx -63.435^{\circ }$.

Part (ix)

Using the reciprocal identity: ${\csc }^{-1}(x)={\sin }^{-1}\left(\frac{1}{x}\right)$. ${\csc }^{-1}(1.172)={\sin }^{-1}\left(\frac{1}{1.172}\right)$ Calculator evaluation: $\theta ={\csc }^{-1}(1.172)\approx 58.566^{\circ }$.

Part (x)

Let $\theta ={\csc }^{-1}\left(\frac{-5}{3}\right)$. This implies $\csc \theta =-\frac{5}{3}$, so $\sin \theta =-\frac{3}{5}$. Here, $P=-3$ and $H=5$. Using the Pythagorean theorem: $\begin{array}{rl}{B}^2& =H^2-P^2\\ B^2& =5^2-(-3{)}^2=25-9=16\\ B& =4\end{array}$ Now, $\tan \theta =\frac{P}{B}=\frac{-3}{4}$. Therefore, ${\csc }^{-1}\left(\frac{-5}{3}\right)={\tan }^{-1}\left(-\frac{3}{4}\right)$. Calculator evaluation: $\theta ={\csc }^{-1}\left(\frac{-5}{3}\right)\approx -36.870^{\circ }$.

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Key Formulas or Methods Used

• Pythagorean Theorem: $H^2=B^2+P^2$
• Trigonometric Ratios:
  • $\sin \theta =\frac{P}{H}$
  • $\cos \theta =\frac{B}{H}$
  • $\tan \theta =\frac{P}{B}$
  • $\cot \theta =\frac{B}{P}$
  • $\sec \theta =\frac{H}{B}$
  • $\csc \theta =\frac{H}{P}$
• Reciprocal Identities: ${\csc }^{-1}(x)={\sin }^{-1}(1/x)$, etc.

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Summary of Steps

1. Set the given inverse trigonometric function equal to $\theta$.
2. Convert the inverse function into a standard trigonometric ratio (e.g., $\cos \theta =x$).
3. Identify the two known sides of the right-angled triangle from the ratio.
4. Calculate the third side using the Pythagorean theorem.
5. Write the new trigonometric ratio for the required function.
6. Use a calculator to find the numerical value of the angle in degrees.