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Question: Find \({{\sin }^{-1}}\left( -\dfrac{1}{2} \right)\)....

Find sin⁡−1(−12){{\sin }^{-1}}\left( -\dfrac{1}{2} \right).

Explanation

Solution

We first find the principal value of x for which sin⁡(x)=−12\sin \left( x \right)=-\dfrac{1}{2}. In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation sin⁡−1(−12){{\sin }^{-1}}\left( -\dfrac{1}{2} \right).

Complete step by step answer:
It’s asked to find sin⁡−1(−12){{\sin }^{-1}}\left( -\dfrac{1}{2} \right). The value in fraction is −12-\dfrac{1}{2}. We need to find x for which sin⁡(x)=−12\sin \left( x \right)=-\dfrac{1}{2}.
We know that in the principal domain or the periodic value of −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} for sin⁡x\sin x, if we get sin⁡a=sin⁡b\sin a=\sin b where −π2≤a,b≤π2-\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} then a=ba=b.
We have the value of sin⁡(−π6)\sin \left( -\dfrac{\pi }{6} \right) as −12-\dfrac{1}{2}. −π2<−π6<π2-\dfrac{\pi }{2}<-\dfrac{\pi }{6}<\dfrac{\pi }{2}.
Therefore, sin⁡(x)=−12=sin⁡(−π6)\sin \left( x \right)=-\dfrac{1}{2}=\sin \left( -\dfrac{\pi }{6} \right) which gives x=−π6x=-\dfrac{\pi }{6}.

For sin⁡(x)=−(12)\sin \left( x \right)=-\left( \dfrac{1}{2} \right), the value of x is x=−π6x=-\dfrac{\pi }{6}.
We also can show the solutions (primary and general) of the equation sin⁡(x)=−(12)\sin \left( x \right)=-\left( \dfrac{1}{2} \right) through a graph. We take y=sin⁡(x)=−(12)y=\sin \left( x \right)=-\left( \dfrac{1}{2} \right). We got two equations: y=sin⁡(x)y=\sin \left( x \right) and y=−(12)y=-\left( \dfrac{1}{2} \right). We place them on the graph and find the solutions as their intersecting points.

We can see the primary solution in the interval −π2≤x≤π2-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} is the point A as x=−π6x=-\dfrac{\pi }{6}. All the other intersecting points of the curve and the line are general solutions.

Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to −∞≤x≤∞-\infty \le x\le \infty . In that case we have to use the formula x=nπ+(−1)nax=n\pi +{{\left( -1 \right)}^{n}}a for sin⁡(x)=sin⁡a\sin \left( x \right)=\sin a where −π2≤a≤π2-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}. For our given problem sin⁡(x)=−(12)\sin \left( x \right)=-\left( \dfrac{1}{2} \right), the general solution will be x=nπ−(−1)nπ6x=n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}. Here n∈Zn\in \mathbb{Z}.