Question

What is $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ ?

Answer 1

Hint: In this, we need to solve the given trigonometric function. The given trigonometric function is $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ and we need to solve the given function. We will start solving the brackets using known trigonometric identities.

Complete step by step solution:
The trigonometric function is $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ and we need to solve the given function.
Now, our first task is to cancel out the ‘ $\arcsin $ ‘function.
We have, $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ ,
Using the half-angle formula $\cos 2A = 1 - 2{\sin ^2}A$ .
Out of three half-angle formulas, we have to use the formula in which the \[\sin \] function is coming, so as to cancel out the $\arcsin $ function.
Here, in the given trigonometric function, $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ , if we compare it with the function $\cos 2A$ , we get, $A = \arcsin \left( {\dfrac{3}{5}} \right)$ .
So, the function becomes, $1 - 2{\sin ^2}\left( {\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ .
Now, ${\sin ^2}A$ can be written as ${\left( {\sin A} \right)^2}$ then we can write the above function as $1 - 2{\left( {\sin \left( {\arcsin \left( {\dfrac{3}{5}} \right)} \right)} \right)^2}$ .
Now, we will cancel out $\sin $ with $\arcsin $ , so the equation becomes,
$1 - 2{\left( {\dfrac{3}{5}} \right)^2}$ ,
Next, we need to solve this equation, so first take a square off $\dfrac{3}{5}$ , then, the equation becomes, $1 - 2\left( {\dfrac{9}{{25}}} \right)$ .
Now, multiply $2$ with $\dfrac{9}{{25}}$ , we get, $1 - \dfrac{{18}}{{25}}$ . Then, we will take LCM of $1$ and $25$ , which is $25$ .
The equation becomes, $\dfrac{{25 - 18}}{{25}}$ , Finally, subtract $18$ from $25$ , which gives, $\dfrac{7}{{25}}$ .
Hence, the value of $\cos \left( {2\arcsin \left( {\dfrac{3}{5}} \right)} \right)$ is $\dfrac{7}{{25}}$ .

Note:
 In the following function, ${\sin ^2}\left( {\arcsin \theta } \right)$ is not equal to $\sin \theta $ , i.e., one $\sin $ is cancelled out by $\arcsin $ .Rather, first, write ${\sin ^2}\left( {\arcsin \theta } \right)$ equivalent to ${\left( {\sin \left( {\arcsin \theta } \right)} \right)^2}$ , then cancel out $\sin $ by $\arcsin $ which is equal to ${\theta ^2}$ .
Although there are three half-angle formulas $\cos 2A$ , we have to choose the function which is suitable for our given question.