Evaluate: $$\sin^{-1} \left( \sin 3\right) $$

Hint:In this question it is given that we have to find the value of $$\sin^{-1} \left( \sin 3\right) $$. So to find the solution we have ro know that,
$$\sin^{-1} \left( \sin \theta \right) =\theta$$ if $$-\dfrac{\pi }{2} \leq \theta \leq \dfrac{\pi }{2}$$..............(1)

Complete step by step answer:
As we know that $$\pi \approx 3.14$$,
Therefore $$\left[ -\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right] =\left[ -1.57,1.57\right] $$
So for $$\sin^{-1} \left( \sin 3\right) $$ 3 lies outside the interval.
Now we have to use a trigonometric formula, i.e, $$\sin \theta =\sin \left( \pi -\theta \right) $$
So by the above formula $$\sin 3$$ can be written as $$\sin \left( \pi -3\right) $$
Now $$\left( \pi -3\right) $$ belongs to the interval $$\left[ -\dfrac{\pi }{2} ,\dfrac{\pi }{2} \right]$$

So by the formula (1) we can write,
$$\sin^{-1} \left( \sin 3\right) =\sin^{-1} \left( \sin \left( \pi -3\right) \right) =\pi -3$$
Therefore the value of $$\sin^{-1} \left( \sin 3\right) $$ is $$\pi -3$$.

Note:
To solve this type of question you need to know that if you don’t know the range of $$\theta$$ then you cannot be able to apply the formula (1). Also in the question in place of $$\theta$$ it is given 3 which is in radian so that is why we used $$\pi$$ as the 3.14(radius value).