Question

How do you find the exact value of $\arcsin 1$ ?

Answer 1

Here in the given question they have asked for $\arcsin 1$ in which arcsin is nothing but the inverse function of the sine. Therefore we need to find the value of the inverse sine function for given value. By referring to the standard values of the trigonometric function we can find the required value.

Complete step-by-step answer:
First we need to know what is $\arcsin 1$ before evaluation.
$\arcsin 1$ is nothing but the inverse sine function of $1$.
In general, the arc sine function is defined in the range of $[0,\pi )$ .
The below table gives the details of the values of trigonometric functions:

{}&0&{\dfrac{\pi }{6}}&{\dfrac{\pi }{4}}&{\dfrac{\pi }{3}}&{\dfrac{\pi }{2}} \\\ {\sin \theta }&0&{\dfrac{1}{2}}&{\dfrac{1}{{\sqrt 2 }}}&{\dfrac{{\sqrt 3 }}{2}}&1 \\\ {\cos \theta }&1&{\dfrac{{\sqrt 3 }}{2}}&{\dfrac{1}{{\sqrt 2 }}}&{\dfrac{1}{2}}&0 \\\ {\tan \theta }&0&{\dfrac{1}{{\sqrt 3 }}}&1&{\sqrt 3 }&\infty \\\ {\csc \theta }&\infty &2&{\sqrt 2 }&{\dfrac{2}{{\sqrt 3 }}}&0 \\\ {\sec \theta }&1&{\dfrac{2}{{\sqrt 3 }}}&{\sqrt 2 }&2&\infty \\\ {\cot \theta }&\infty &{\sqrt 3 }&1&{\dfrac{1}{{\sqrt 3 }}}&0 \end{array}$$ By referring to these trigonometric values of standard angle we can solve the problems easily. Before finding the value for sine inverse function, first we find the sine function value then we do inverse, only because to understand the concept better. From the above table we notice that $\sin \theta = 1$ when $\theta = {90^ \circ }$ or $\theta = \dfrac{\pi }{2}$ . (if we transfer the sine function from left hand side to right hand side we get inverse sine function) Now we can find the value for $\arcsin 1$ that is ${\sin ^{ - 1}}1$. Therefore by referring to the above discussion $\arcsin 1 = \dfrac{\pi }{2}$ or ${\sin ^{ - 1}}1 = \dfrac{\pi }{2}$ . Or we can also reduce $\sin \theta = 1$ when $\theta = \dfrac{\pi }{2}$ as $\sin \dfrac{\pi }{2} = 1$ $ \Rightarrow \dfrac{\pi }{2} = \dfrac{1}{{\sin }}$ which can be rewrite as $ \Rightarrow \dfrac{\pi }{2} = {\sin ^{ - 1}}1$ **Hence we can say ${\sin ^{ - 1}}1 = \dfrac{\pi }{2}$ .** **Note:** By using the predefined values of trigonometric functions we can arrive at the correct answer easily. One thing we need to remember here is values if you don’t know the values of standard functions then it is difficult to solve the problems.