How is cos^-1(3/5) related to this θ =…

How is cos^-1(3/5) related to this θ = cos^-1(-3/5)?

Question 116

HardOriginal question

Let $\theta = \cos^{-1}\left(-\frac{3}{5}\right)$ .

How is $\cos^{-1}\left(\frac{3}{5}\right)$ related to this $\theta = \cos^{-1}\left(-\frac{3}{5}\right)$ ?

\cos^{-1}\left(\frac{3}{5}\right) = \pi - \theta

Correct answerYour answer

\cos^{-1}\left(\frac{3}{5}\right) = -\theta

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\cos^{-1}\left(\frac{3}{5}\right) = \theta - \frac{\pi}{2}

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\cos^{-1}\left(\frac{3}{5}\right) = \frac{\pi}{2} - \theta

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\cos^{-1}(-x) = \pi - \cos^{-1} x$ , so $\theta = \pi - \cos^{-1}\left(\frac{3}{5}\right)$ .
Hence $\cos^{-1}\left(\frac{3}{5}\right) = \pi - \theta$ .

Worth noting

$-\theta$ is the rule for $\sin^{-1}$ , not $\cos^{-1}$ .

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.