Question

How do you evaluate $\arcsin \left( \dfrac{4}{5} \right)$ ?

Answer 1

Hint: The arcsin(x) is a mathematical function which is equal to the inverse of sine function. Suppose we have a function $y=f(x)$. Then the inverse of the function $f(x)$ is a function in which for a value of y we have a value of x.

Complete step by step solution:
Let us first understand what is meant by arcsin(x).The arcsin(x) is a mathematical function which is equal to the inverse of sine function.i.e. $\arcsin (x)={{\sin }^{-1}}(x)$. Suppose we have a function $y=f(x)$. Then the inverse of the function $f(x)$ is a function in which for a value of y we have a value of x. In other words, the inverse of a function is $x={{f}^{-1}}(y)$.

Therefore, if we have a function $y={{f}^{-1}}(x)$, then we can write that $f(y)=x$.
In the given case, $y={{f}^{-1}}(x)={{\sin }^{-1}}(x)$.
Then, this means that $\sin (y)=x$.
Since we have to calculate the value of $\arcsin \left( \dfrac{4}{5} \right)$, the value of x is $\dfrac{4}{5}$.
This means that $\sin (y)=\dfrac{4}{5}$.
Therefore, we have to find that angle for which sine of that angle is equal to $\dfrac{4}{5}$.
And we know that $\sin {{53}^{\circ }}=\dfrac{4}{5}$.
This means that $y={{53}^{\circ }}$.
But, we know that $y={{\sin }^{-1}}(x)$
${{\sin }^{-1}}(x)=\arcsin (x)$.
$\Rightarrow y=\arcsin \left( \dfrac{4}{5} \right)\\
\therefore y={{53}^{\circ }}$

Note:If you do not know that $\sin {{53}^{\circ }}=\dfrac{4}{5}$, then you can calculate it by using trigonometry if you know $\tan {{53}^{\circ }}=\dfrac{4}{3}$. We know that tangent of an angle (say x) is given to be equal to the ratio of the opposite side to that angle to the adjacent side to that angle of the right angled triangle of whose angle is.
i.e. $\tan x=\dfrac{\text{Opposite}}{\text{Adjacent}}$. In this case, $x={{53}^{\circ }}$ and we know that $\tan {{53}^{\circ }}=\dfrac{4}{3}$
Therefore,
$\tan {{53}^{\circ }}=\dfrac{\text{Opposite}}{\text{Adjacent}}=\dfrac{4}{3}$
Now, draw a right angled triangle, whose one angle is of $x={{53}^{\circ }}$ , with the opposite side to this angle of length 4 units and the adjacent side of length 3 units.
Then by Pythagoras theorem we know that ${{\text{(hypotenuse)}}^{2}}={{3}^{2}}+{{4}^{2}}=25$
Therefore, $\text{hypotenuse}=5$
Now, we can use the relation that $\sin x=\sin {{53}^{\circ }}=\dfrac{\text{Opposite}}{\text{Hypotenuse}}$.
Then this means that $\sin {{53}^{\circ }}=\dfrac{4}{5}$