Question

How do you evaluate $\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$ without a calculator?

Answer 1

Hint: Here without using a calculator also, we can solve it easily. We know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ and therefore we can get the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ and then we can put that angle to find the value of $\sin $ function of that angle.

Complete step by step solution:
Here we are given to solve the trigonometric function which is given as $\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$
Now we need to solve it without using a calculator. Hence we can solve it by taking the general values of the trigonometric functions. We must know that:
$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$$ - - - - - - (1)$
Whenever we are given this type of function where we are given inverse of any function, we can take it to other side and remove the inverse and write it as:
$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$$ = \dfrac{\pi }{3}$$ - - - - (2)$
Now we know that we are given to find the value of $\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$
Now substituting the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$$ = \dfrac{\pi }{3}$ in the above given problem we will get that:
$\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$
$\sin \left( {\dfrac{\pi }{3}} \right)$
Now we know that $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Hence we have solved this without using the calculator and we get the value of

$\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$$ = \dfrac{{\sqrt 3 }}{2}$

Hence we come to know that for such problems we must know the general values of the trigonometric functions as this will be very helpful to solve such problems.

Note:
Here we can also do it be other method where we can let ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x$
So we can say $\cos x = \dfrac{1}{2}$
Now we know that
$
  {\sin ^2}x + {\cos ^2}x = 1 \\
  {\sin ^2}x = 1 - {\cos ^2}x \\
 $
Hence we can say that
$\sin x = \sqrt {1 - {{\cos }^2}x} $
Substituting this value we will get:
$\sin x = \sqrt {1 - {{\cos }^2}x} = \sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} = \dfrac{{\sqrt 3 }}{2}$
Hence we get that:
$x = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Hence we get:
$\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$$ = \dfrac{{\sqrt 3 }}{2}$
Hence in this way also we can solve the problem in a better way.