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Inverse Trigonometric Functions · Question 79 of 120

Question 79

MediumOriginal question

Read the two statements and choose the correct option.

I. The curves y=sin⁡−1xy = \sin^{-1} x and y=cos⁡−1xy = \cos^{-1} x are mirror images of each other in the line y=xy = x.

II. The curves y=tan⁡−1xy = \tan^{-1} x and y=cot⁡−1xy = \cot^{-1} x are mirror images of each other in the line y=π4y = \frac{\pi}{4}.

Options for question 79

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. I: the mirror image of y=sin⁡−1xy = \sin^{-1} x in y=xy = x is y=sin⁡xy = \sin x, −π2≤x≤π2-\frac{\pi}{2} \le x \le \frac{\pi}{2}, which runs over [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], not [−1,1][-1, 1]. False.

  2. II: cot⁡−1x=π2−tan⁡−1x\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x, so at each xx the two heights are equally far above and below π4\frac{\pi}{4}. True.

Worth noting

Reflecting in y=xy = x turns a curve into the curve of its inverse, and cos⁡−1\cos^{-1} is not the inverse of sin⁡−1\sin^{-1}.

Concept
Reflections and symmetry of the curves
Typical time
90 s
Your time
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The rule this testsGraphs of the inverse functions
Practise this chapter →

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

Practise this chapter →

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