Question

How do you evaluate \[\sin \left( {\arccos \left( {\dfrac{3}{5}} \right)} \right)\] ?

Answer 1

Hint: We are about to evaluate the given identities combination. We know that arc stands for inverse of a function. So we will equate the identity in the bracket to an angle such that the inverse of the function will give a value.

Complete step by step solution:
Given that \[\sin \left( {\arccos \left( {\dfrac{3}{5}} \right)} \right)\].
Let \[\arccos \left( {\dfrac{3}{5}} \right) = A\]
So we can rewrite the above term as
\[co{s^{ - 1}}\left( {\dfrac{3}{5}} \right) = A\]
Taking the trigonometric function on one side we get,
\[\cos A = \dfrac{3}{5}\]
This is the value of one function. Now we will get to know the value of sin,
We know that \[{\sin ^2}x = 1 - {\cos ^2}x\]
This can be written as
\[\sin x = \sqrt {1 - {{\cos }^2}x} \]
In our case above is
\[\sin A = \sqrt {1 - {{\cos }^2}A} \]
Putting the value of cos function
\[\sin A = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} \]
Taking the square
\[\sin A = \sqrt {1 - \left( {\dfrac{9}{{25}}} \right)} \]
Taking LCM
\[\sin A = \sqrt {\dfrac{{25 - 9}}{{25}}} \]
\[\Rightarrow\sin A = \sqrt {\dfrac{{16}}{{25}}} \]
Under root value is the perfect square so taking the root
\[\therefore\sin A = \dfrac{4}{5}\]

So we will get the answer as \[\sin \left( {\arccos \left( {\dfrac{3}{5}} \right)} \right) = \dfrac{4}{5}\].

Note: We are about to find the value of sin function so we used arc function as a substitution and used standard identities. Also note that we have received the value of sin directly from the formula so no need to write that again. Remember in trigonometric problems generally we require the different identities to solve them more efficiently.