Question

How do you find the exact solution to $\arcsin \left( \dfrac{\sqrt{3}}{2} \right)$ ?

Answer 1

Hint: We are given that $arc\left[ \sin \left( \dfrac{\sqrt{3}}{2} \right) \right]$ , we have to find solution, we start by understanding what arcsin stand for, then we will start solution by considering ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ as $\theta $ , then we multiply both side by sin, then we use relation $\sin \left( {{\sin }^{-1}}\theta \right)=\theta $ to reduce the equation, we will also use the concept that sin is a periodic function, and lastly using the information that sin is positive in Quad I and II we will solve our problem.

Complete step by step answer:
We are given $arc\left( \sin \left( \dfrac{\sqrt{3}}{2} \right) \right)$ , here are sin stand for ${{\sin }^{-1}}$ so, $arc\left( \sin \left( \dfrac{\sqrt{3}}{2} \right) \right)$ mean ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ .
Now we consider ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ as \[\theta \] , so
${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\theta $ …………………………………… (1)
Now we will use identity given as –
$\sin \left( {{\sin }^{-1}}\left( \theta \right) \right)=\theta $
To simplify above given equation –
We have to find value of ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ mean we have to find value of $\theta $ .
We multiply equation (1) by sin on both side, we get –
$\sin \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)=\sin \theta $
As $\sin \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)=\dfrac{\sqrt{3}}{2}$
$\Rightarrow \dfrac{\sqrt{3}}{2}=\sin \theta $
Now, we get $\sin \theta =\dfrac{\sqrt{3}}{2}$ , it is positive, the sin take positive value is quadrant I and II.
So, $\theta $ will be in Quadrant I and II.
In Quadrant I, we know –
$\sin \theta =\dfrac{\sqrt{3}}{2}$ is true.
For $\theta =\dfrac{\pi }{3}\left( {{60}^{\circ }} \right)$
Hence the solution is $\theta =\dfrac{\pi }{3}$ .
As we know sin is periodic, so it will repeat the value of every $2\pi $ period.
So the general solution is $x=\dfrac{\pi }{3}+2n\pi ,n\in N$ .
 For Quadrant II,
The solution for $\sin \theta =\dfrac{\sqrt{3}}{2}$ is given as –
$x=\pi -\theta $ , where $\theta $ is the solution of $\sin \theta =\dfrac{\sqrt{3}}{2}$ in Quadrant I.
In Quadrant I, we get $\theta $ as $\dfrac{\pi }{3}$ so, solution is Quadrant II in $x=\pi -\dfrac{\pi }{3}$ ,
By simplifying, we get –
$x=\dfrac{2\pi }{3}$ .
As again sin is periodic so it will repeat it value after $2\pi $ so, general solution is given as –
$x=\dfrac{2\pi }{3}+2n\pi $ , where $n\in N$ .
Hence we get –
Solution of ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ are –
$x=\dfrac{2\pi }{3}+2n\pi $ and $x=\dfrac{\pi }{3}+2n\pi ,n\in N$

Note: While solving such problems we should be very careful with identity like $\sin 2x\ne 2\sin x$ or $\cos 2\theta \ne 2\cos \theta \sin \theta $ we should not mix or use appropriate identity. Also we should always cross check solutions so that the chance of error will get eliminated.
While solving fractions we always report answers in the simplest form so if something is common we need to cancel it always.
Remember if we are not given the domain of the function so we will always give a general solution, that stays true for all domains if we solve $\sin x=1$ .
And write the solution $x=\dfrac{\pi }{2}$ then it will not be fully correct.
So, we need to be clear when we need to write the general solution.