Why isn't just equal to 8?

The range of the principal branch of is . Since is far outside this interval, we must find an angle such that . Since , , and , so .

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Trigonometry: JEE Main Mathematics Question with Solution

\sin^{-1} \frac{\alpha}{17} + \cos^{-1} \frac{4}{5} - \tan^{-1} \frac{77}{36} = 0, 0 < \alpha < 13

, then

\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)

is equal to

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A16B0C

\pi

16 - 5\pi

Step-by-step solutionView

Correct answer

Solving the inverse trigonometric equation yields $\alpha = 8$ , for which $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi$ .

Option analysis

Why each option works or fails

A · 16

Assumed $\sin^{-1}(\sin \alpha) = \alpha$ and $\cos^{-1}(\cos \alpha) = \alpha$ without checking principal value ranges, yielding $\alpha + \alpha = 2(8) = 16$ . Check the principal value branch: $\sin^{-1}(\sin x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos^{-1}(\cos x) \in [0, \pi]$ . Since $8$ lies outside these intervals, reduce using periodicity and symmetry.

B · 0

Confused the signs in the branch reduction formulas, writing $\sin^{-1}(\sin 8) = 8 - 3\pi$ instead of $3\pi - 8$ , leading to $(8 - 3\pi) + (8 - 2\pi) = 16 - 5\pi$ or miscanceling to $0$ . Ensure the reduced value falls strictly within the principal branch: since $8 \approx 2.55\pi$ , $3\pi - 8 \approx 1.42 \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ , whereas $8 - 3\pi$ is negative and outside.

C ·

\pi

None. The value $\alpha = 8$ is correctly found and evaluated on the proper branches of $\sin^{-1}$ and $\cos^{-1}$ . None.

D ·

16 - 5\pi

Inverted the branch for $\sin^{-1}(\sin 8)$ as $8 - 3\pi$ instead of $3\pi - 8$ , resulting in $(8 - 3\pi) + (8 - 2\pi) = 16 - 5\pi$ . Remember that $\sin^{-1}(\sin x) = \pi - x$ or $3\pi - x$ around odd multiples of $\pi$ to maintain the correct slope and range.

Reviewed route

Solution

StepWorking

01given

We are given $\sin^{-1}\left(\frac{\alpha}{17}\right) + \cos^{-1}\left(\frac{4}{5}\right) - \tan^{-1}\left(\frac{77}{36}\right) = 0$ with $0 < \alpha < 13$ .

02goal

Find the value of $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ .

03approach

Convert $\cos^{-1}\left(\frac{4}{5}\right)$ into $\tan^{-1}$ using right-triangle relations, rearrange the equation to isolate $\sin^{-1}\left(\frac{\alpha}{17}\right)$ , apply the subtraction formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ , solve for $\alpha$ , and finally compute $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ taking the principal branch intervals into account.

04execute

Convert $\cos^{-1}\left(\frac{4}{5}\right)$ to $\tan^{-1}$ : since $\cos \theta = \frac{4}{5}$ , $\tan \theta = \frac{3}{4}$ , so $\cos^{-1}\left(\frac{4}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right)$ . Rearrange: $\sin^{-1}\left(\frac{\alpha}{17}\right) = \tan^{-1}\left(\frac{77}{36}\right) - \tan^{-1}\left(\frac{3}{4}\right)$ Apply subtraction formula: $\tan^{-1}\left(\frac{77}{36}\right) - \tan^{-1}\left(\frac{3}{4}\right) = \tan^{-1}\left(\frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}}\right) = \tan^{-1}\left(\frac{\frac{77 - 27}{36}}{\frac{144 + 231}{144}}\right) = \tan^{-1}\left(\frac{50}{36} \cdot \frac{144}{375}\right) = \tan^{-1}\left(\frac{25}{18} \cdot \frac{48}{125}\right) = \tan^{-1}\left(\frac{8}{15}\right)$ Thus, $\sin^{-1}\left(\frac{\alpha}{17}\right) = \tan^{-1}\left(\frac{8}{15}\right)$ . Since $\tan \phi = \frac{8}{15}$ , the hypotenuse is $\sqrt{8^2 + 15^2} = 17$ , so $\sin \phi = \frac{8}{17}$ . Therefore, $\frac{\alpha}{17} = \frac{8}{17} \implies \alpha = 8$ , which satisfies $0 < \alpha < 13$ .

05execute

Now evaluate $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8)$ : Note that $2\pi \approx 6.283$ and $\frac{5\pi}{2} \approx 7.854$ , $3\pi \approx 9.425$ . Since $8 \in \left(\frac{5\pi}{2}, 3\pi\right)$ : $\sin^{-1}(\sin 8) = 3\pi - 8$ . For $\cos^{-1}(\cos 8)$ , since $8 \in [2\pi, 3\pi]$ : $\cos^{-1}(\cos 8) = 8 - 2\pi$ . Summing both expressions: $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi.$

✓verify

Check: $3\pi - 8 \approx 9.425 - 8 = 1.425 \in [-\pi/2, \pi/2]$ since $\pi/2 \approx 1.571$ . Thus the principal branch reduction for $\sin^{-1}$ is correct. Check: $8 - 2\pi \approx 8 - 6.283 = 1.717 \in [0, \pi]$ since $\pi \approx 3.142$ . Thus the principal branch reduction for $\cos^{-1}$ is correct. Their sum is $(3\pi - 8) + (8 - 2\pi) = \pi$ .

Hints that build this answer step by step

How can the equation $\sin^{-1} \frac{\alpha}{17} + \cos^{-1} \frac{4}{5} = \tan^{-1} \frac{77}{36}$ be rewritten in terms of a single inverse function to solve for $\alpha$ ?
Convert all terms to $\tan^{-1}$ so that $\tan^{-1}\left(\frac{\alpha}{\sqrt{289-\alpha^2}}\right) = \tan^{-1}\frac{77}{36} - \tan^{-1}\frac{3}{4}$
What is the value of $\tan^{-1}\frac{77}{36} - \tan^{-1}\frac{3}{4}$ ?
$\tan^{-1}\left(\frac{8}{15}\right)$
Given $\tan^{-1}\left(\frac{\alpha}{\sqrt{289-\alpha^2}}\right) = \tan^{-1}\left(\frac{8}{15}\right)$ , what is $\alpha$ ?
$\alpha = 8$
Since $\alpha = 8$ and $2.5\pi < 8 < 3\pi$ (specifically, $\frac{5\pi}{2} \approx 7.85$ and $3\pi \approx 9.42$ ), what are $\sin^{-1}(\sin 8)$ and $\cos^{-1}(\cos 8)$ ?
$\sin^{-1}(\sin 8) = 3\pi - 8$ and $\cos^{-1}(\cos 8) = 8 - 2\pi$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why isn't

\sin^{-1}(\sin 8)

just equal to 8?

The range of the principal branch of $\sin^{-1} x$ is $[-\pi/2, \pi/2] \approx [-1.57, 1.57]$ . Since $8$ is far outside this interval, we must find an angle $\theta \in [-\pi/2, \pi/2]$ such that $\sin \theta = \sin 8$ . Since $8 \in (5\pi/2, 3\pi)$ , $\sin(3\pi - 8) = \sin 8$ , and $3\pi - 8 \approx 1.425 \in [-\pi/2, \pi/2]$ , so $\sin^{-1}(\sin 8) = 3\pi - 8$ .

Answer

Solving the inverse trigonometric equation yields $\alpha = 8$ , for which $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi$ .

Why each option works or fails

A: 16 - Assumed $\sin^{-1}(\sin \alpha) = \alpha$ and $\cos^{-1}(\cos \alpha) = \alpha$ without checking principal value ranges, yielding $\alpha + \alpha = 2(8) = 16$ . Check the principal value branch: $\sin^{-1}(\sin x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos^{-1}(\cos x) \in [0, \pi]$ . Since $8$ lies outside these intervals, reduce using periodicity and symmetry.
B: 0 - Confused the signs in the branch reduction formulas, writing $\sin^{-1}(\sin 8) = 8 - 3\pi$ instead of $3\pi - 8$ , leading to $(8 - 3\pi) + (8 - 2\pi) = 16 - 5\pi$ or miscanceling to $0$ . Ensure the reduced value falls strictly within the principal branch: since $8 \approx 2.55\pi$ , $3\pi - 8 \approx 1.42 \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ , whereas $8 - 3\pi$ is negative and outside.
C · correct: $\pi$ - None. The value $\alpha = 8$ is correctly found and evaluated on the proper branches of $\sin^{-1}$ and $\cos^{-1}$ . None.
D: $16 - 5\pi$ - Inverted the branch for $\sin^{-1}(\sin 8)$ as $8 - 3\pi$ instead of $3\pi - 8$ , resulting in $(8 - 3\pi) + (8 - 2\pi) = 16 - 5\pi$ . Remember that $\sin^{-1}(\sin x) = \pi - x$ or $3\pi - x$ around odd multiples of $\pi$ to maintain the correct slope and range.

Step-by-step solution

given: We are given $\sin^{-1}\left(\frac{\alpha}{17}\right) + \cos^{-1}\left(\frac{4}{5}\right) - \tan^{-1}\left(\frac{77}{36}\right) = 0$ with $0 < \alpha < 13$ .
goal: Find the value of $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ .
approach: Convert $\cos^{-1}\left(\frac{4}{5}\right)$ into $\tan^{-1}$ using right-triangle relations, rearrange the equation to isolate $\sin^{-1}\left(\frac{\alpha}{17}\right)$ , apply the subtraction formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ , solve for $\alpha$ , and finally compute $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ taking the principal branch intervals into account.
execute: Convert $\cos^{-1}\left(\frac{4}{5}\right)$ to $\tan^{-1}$ : since $\cos \theta = \frac{4}{5}$ , $\tan \theta = \frac{3}{4}$ , so $\cos^{-1}\left(\frac{4}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right)$ . Rearrange: $\sin^{-1}\left(\frac{\alpha}{17}\right) = \tan^{-1}\left(\frac{77}{36}\right) - \tan^{-1}\left(\frac{3}{4}\right)$ Apply subtraction formula: $\tan^{-1}\left(\frac{77}{36}\right) - \tan^{-1}\left(\frac{3}{4}\right) = \tan^{-1}\left(\frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}}\right) = \tan^{-1}\left(\frac{\frac{77 - 27}{36}}{\frac{144 + 231}{144}}\right) = \tan^{-1}\left(\frac{50}{36} \cdot \frac{144}{375}\right) = \tan^{-1}\left(\frac{25}{18} \cdot \frac{48}{125}\right) = \tan^{-1}\left(\frac{8}{15}\right)$ Thus, $\sin^{-1}\left(\frac{\alpha}{17}\right) = \tan^{-1}\left(\frac{8}{15}\right)$ . Since $\tan \phi = \frac{8}{15}$ , the hypotenuse is $\sqrt{8^2 + 15^2} = 17$ , so $\sin \phi = \frac{8}{17}$ . Therefore, $\frac{\alpha}{17} = \frac{8}{17} \implies \alpha = 8$ , which satisfies $0 < \alpha < 13$ .
execute: Now evaluate $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8)$ : Note that $2\pi \approx 6.283$ and $\frac{5\pi}{2} \approx 7.854$ , $3\pi \approx 9.425$ . Since $8 \in \left(\frac{5\pi}{2}, 3\pi\right)$ : $\sin^{-1}(\sin 8) = 3\pi - 8$ . For $\cos^{-1}(\cos 8)$ , since $8 \in [2\pi, 3\pi]$ : $\cos^{-1}(\cos 8) = 8 - 2\pi$ . Summing both expressions: $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi.$
verify: Check: $3\pi - 8 \approx 9.425 - 8 = 1.425 \in [-\pi/2, \pi/2]$ since $\pi/2 \approx 1.571$ . Thus the principal branch reduction for $\sin^{-1}$ is correct. Check: $8 - 2\pi \approx 8 - 6.283 = 1.717 \in [0, \pi]$ since $\pi \approx 3.142$ . Thus the principal branch reduction for $\cos^{-1}$ is correct. Their sum is $(3\pi - 8) + (8 - 2\pi) = \pi$ .

Trigonometry: JEE Main Mathematics Question with Solution

What feels right?

Solving the inverse trigonometric equation yields $\alpha = 8$ , for which $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Trigonometry: JEE Main Mathematics Question with Solution

Solving the inverse trigonometric equation yields α=8\alpha = 8α=8, for which sin⁡−1(sin⁡8)+cos⁡−1(cos⁡8)=(3π−8)+(8−2π)=π\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pisin−1(sin8)+cos−1(cos8)=(3π−8)+(8−2π)=π.

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03approach

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Solving the inverse trigonometric equation yields $\alpha = 8$ , for which $\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi$ .

Let $(a, b) \subset (0, 2\pi)$ be the largest interval for which…