Question

What is the Arcsine of $\dfrac{3}{5}$?

Answer 1

Hint: We solve this problem by using the standard value of angle that is,
$\sin {{37}^{\circ }}=\dfrac{3}{5}$
By using this standard value we find the principle solution and then we use the standard conversion of sine ratio which is given as,
$\sin \left( \left( 2n+1 \right)\dfrac{\pi }{2}\pm \theta \right)=\sin \theta ;n=0,2,4,6,.....$
By using this conversion we find the general solution of the required value.

Complete step-by-step answer:
We are asked to find the arcsine of $\dfrac{3}{5}$
Let us assume that the required answer as,
$\Rightarrow x=\arcsin \left( \dfrac{3}{5} \right)$
Now, let us apply sine ratio to both sides then we get,
$\begin{align}
  & \Rightarrow \sin x=\sin \left( \arcsin \left( \dfrac{3}{5} \right) \right) \\
 & \Rightarrow \sin x=\dfrac{3}{5}..........equation(i) \\
\end{align}$
We know that the value of standard angle that is,
$\sin {{37}^{\circ }}=\dfrac{3}{5}..............equation(ii)$

Now, by comparing both equation (i) and equation (ii) we get the principle value of $'x'$ as,
$\Rightarrow x={{37}^{\circ }}$
Now, let us find the general solution of that equation.
We know that the standard conversion of sine ratio which is given as,
$\sin \left( \left( 2n+1 \right)\dfrac{\pi }{2}\pm \theta \right)=\sin \theta ;n=0,2,4,6,.....$
Now let us assume $\theta =x$ then we get,
$\Rightarrow \sin \left( \left( 2n+1 \right)\dfrac{\pi }{2}\pm x \right)=\sin x=\dfrac{3}{5}..........equation(iii)$ where, $n=0,2,4,6,.....$
Now, by comparing both equation (ii) and equation (iii) we get the required answer as,
$\begin{align}
  & \Rightarrow \left( \left( 2n+1 \right)\dfrac{\pi }{2}\pm x \right)={{37}^{\circ }} \\
 & \Rightarrow \pm x=-\left( 2n+1 \right)\dfrac{\pi }{2}+{{37}^{\circ }} \\
\end{align}$
By taking the $'\pm '$ sign from LHS to RHS then we get,
\[\Rightarrow x=\mp \left( 2n+1 \right)\dfrac{\pi }{2}\pm {{37}^{\circ }}\]
Here, we can divide the above result into two solutions as,
$\begin{align}
  & \Rightarrow x=-\left( 2n+1 \right)\dfrac{\pi }{2}+{{37}^{\circ }} \\
 & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}-{{37}^{\circ }} \\
\end{align}$
Where, $n=0,2,4,6,.....$
Therefore, we can conclude that the general solution of arcsine of $\dfrac{3}{5}$ is given as,
$\begin{align}
  & \therefore \arcsin \left( \dfrac{3}{5} \right)=-\left( 2n+1 \right)\dfrac{\pi }{2}+{{37}^{\circ }} \\
 & \text{OR }\arcsin \left( \dfrac{3}{5} \right)=\left( 2n+1 \right)\dfrac{\pi }{2}-{{37}^{\circ }} \\
\end{align}$
Where, $n=0,2,4,6,.....$

Note: The main mistake that can be done is not giving the answer as a general solution. We directly know that the arcsine of $\dfrac{3}{5}$ is ${{37}^{\circ }}$ which will be the principal solution.
But we need to give the solution of inverse trigonometric ratios always in general solution. Because the trigonometric ratios are all periodic functions which says that for many values of angles the value of trigonometric ratios of those angles gives the same answer.
In reverse for inverse ratios of values there will be many angles. So, we need to generalise all the values of angles and give the general solution rather than principle solution.