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

How do you find the exact solution to arcsin⁡(32)\arcsin \left( \dfrac{\sqrt{3}}{2} \right) ?

Explanation

Solution

We are given that arc[sin⁡(32)]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(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) as θ\theta , then we multiply both side by sin, then we use relation sin⁡(sin⁡−1θ)=θ\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(sin⁡(32))arc\left( \sin \left( \dfrac{\sqrt{3}}{2} \right) \right) , here are sin stand for sin⁡−1{{\sin }^{-1}} so, arc(sin⁡(32))arc\left( \sin \left( \dfrac{\sqrt{3}}{2} \right) \right) mean sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) .
Now we consider sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) as θ\theta , so
sin⁡−1(32)=θ{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\theta …………………………………… (1)
Now we will use identity given as –
sin⁡(sin⁡−1(θ))=θ\sin \left( {{\sin }^{-1}}\left( \theta \right) \right)=\theta
To simplify above given equation –
We have to find value of sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) mean we have to find value of θ\theta .
We multiply equation (1) by sin on both side, we get –
sin⁡(sin⁡−1(32))=sin⁡θ\sin \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)=\sin \theta
As sin⁡(sin⁡−1(32))=32\sin \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \right)=\dfrac{\sqrt{3}}{2}
⇒32=sin⁡θ\Rightarrow \dfrac{\sqrt{3}}{2}=\sin \theta
Now, we get sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} , it is positive, the sin take positive value is quadrant I and II.
So, θ\theta will be in Quadrant I and II.
In Quadrant I, we know –
sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} is true.
For θ=π3(60∘)\theta =\dfrac{\pi }{3}\left( {{60}^{\circ }} \right)
Hence the solution is θ=π3\theta =\dfrac{\pi }{3} .
As we know sin is periodic, so it will repeat the value of every 2π2\pi period.
So the general solution is x=π3+2nπ,n∈Nx=\dfrac{\pi }{3}+2n\pi ,n\in N .
For Quadrant II,
The solution for sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} is given as –
x=π−θx=\pi -\theta , where θ\theta is the solution of sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} in Quadrant I.
In Quadrant I, we get θ\theta as π3\dfrac{\pi }{3} so, solution is Quadrant II in x=π−π3x=\pi -\dfrac{\pi }{3} ,
By simplifying, we get –
x=2π3x=\dfrac{2\pi }{3} .
As again sin is periodic so it will repeat it value after 2π2\pi so, general solution is given as –
x=2π3+2nπx=\dfrac{2\pi }{3}+2n\pi , where n∈Nn\in N .
Hence we get –
Solution of sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) are –
x=2π3+2nπx=\dfrac{2\pi }{3}+2n\pi and x=π3+2nπ,n∈Nx=\dfrac{\pi }{3}+2n\pi ,n\in N

Note: While solving such problems we should be very careful with identity like sin⁡2x≠2sin⁡x\sin 2x\ne 2\sin x or cos⁡2θ≠2cos⁡θsin⁡θ\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\sin x=1 .
And write the solution x=π2x=\dfrac{\pi }{2} then it will not be fully correct.
So, we need to be clear when we need to write the general solution.