Why is $\sin^{-1}(\sin(\frac{5\pi}4))$ equal to $-\frac{\pi}{4}$ instead of $\frac{\pi}{4}$?

Question

I have the problem of $\arcsin(\sin(\frac{5\pi}4))$. I went to solve this by first finding the reference angle because $\frac{5\pi}4$ is not in the domain of $\arcsin(\sin(x))$. since $\frac{5\pi}4$ is in quadrant III, that would be $\frac{5\pi}4 - \pi$, which is equal to $\frac{\pi}{4}$. Since it is in the domain of $\arcsin(\sin(x))$, then the answer is $\frac{\pi}{4}$. However, all the online calculators say the answer is $-\frac{\pi}{4}$ could someone explain please. Thanks

Answer 1

If you draw the situation geometrically, then yes, as you have pointed out, the reference angle is $\dfrac{5 \pi}{4} - \pi = \dfrac{\pi}{4}$. However, the angle $\dfrac{5 \pi}{4}$ is in quadrant III, as you said. So if you follow the ASTC rule (or whatever rule/mnemonic you know), then you know in quadrant III, $\sin$ is negative. Hence, \begin{align} \sin \left(\frac{5 \pi}{4} \right) = - \sin \left( \frac{\pi}{4} \right) = \sin \left(- \frac{\pi}{4} \right) \tag{$*$} \end{align}

Also, take note that the equation \begin{align} \arcsin(\sin(x)) = x \tag{$**$} \end{align} is true if and only if $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$; this is because of the principal branch of the inverse trigonometric functions. Hence, the answer to your question is \begin{align} \arcsin\left( \sin \left(\frac{5 \pi}{4} \right) \right) &= \arcsin\left( \sin \left(-\frac{ \pi}{4} \right) \right) \tag{by ($*$)} \\ &= -\dfrac{\pi}{4} \tag{by ($**$)} \end{align}

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Alternatively, you can use the fact $\arcsin$ is an odd function, which means we can pull minus signs out: $\arcsin(-\xi) = - \arcsin(\xi)$, for every $\xi$ which satisfies $-1 \leq \xi \leq 1$ (the domain of the $\arcsin$ function). This implies \begin{align} \arcsin\left( \sin \left(\frac{5 \pi}{4} \right) \right) &= \arcsin\left( -\sin \left(\frac{ \pi}{4} \right) \right) \\ &= -\arcsin\left( \sin \left(\frac{ \pi}{4} \right) \right) \\ &= - \dfrac{\pi}{4} \end{align}

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In either case, the thing you missed out on is the fact that in third quadrant $\sin$ is negative. So $\sin(\theta) = - \sin(\text{reference angle of $\theta$})$, if $\theta$ is in quadrant III or IV.

Answer 2

$\sin^{-1}\big(\sin(\frac{5\pi}{4})\big) =\sin^{-1}\big(\sin(\pi+\frac{\pi}{4})\big) = \sin^{-1}(-\sin\frac{\pi}{4}) = -\sin^{-1}(\sin\frac{\pi}{4}) = -\frac{\pi}{4}$

[As $\sin(\pi+x) = -\sin(x) $ and $\sin^{-1}(-x) = -\sin^{-1}(x)$]

Answer 3

All it is is that different calculators use slight variations of the bounded domain for inverse trigonometric functions. The chances are that most online calculators actually just call some atan() or atan2() function from a maths API in Javascript, Python, PHP etc.

In terms of correctness, it really isn't all too important on the end result you get, all that is important is that whatever the domain for your inverse tangent is, ensure that that is entirely what you work with throughout your answers.