Question

If $$\cos (\sin x) = \dfrac{1}{{\sqrt 2 }}$$ , then x must lie in the interval
$\eqalign{
  & 1)\left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)or\left( {\dfrac{\pi }{2},\pi } \right) \cr
  & 2)\left( {\dfrac{{ - \pi }}{4},0} \right) \cr
  & 3)\left( {\pi ,\dfrac{{3\pi }}{2}} \right) \cr
  & 4)\left( {\dfrac{{3\pi }}{2},2\pi } \right) \cr} $

Answer 1

Hint: Since there are two trigonometric functions on the LHS, let’s simplify it down to only one term. The RHS value is $\dfrac{1}{{\sqrt 2 }}$ so let us find out which values will go with it so that we can multiply that trigonometric function on both sides. Then, only one will remain. We can then figure out the range for the function.

Complete step-by-step answer:
The given equation is,
$$\cos (\sin x) = \dfrac{1}{{\sqrt 2 }}$$
Now, let us multiply by ${\cos ^{ - 1}}$ on both sides. We get,
${\cos ^{ - 1}}\left[ {\cos \left( {\sin x} \right)} \right] = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
We know that, ${\cos ^{ - 1}}$and $\cos $ get cancelled out.
Then, we will only be left with the $\sin $ function on the LHS.
$\sin x = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
We know that the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$ is $\dfrac{\pi }{4},\dfrac{{3\pi }}{4}$
Therefore,
$\sin x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4}$
Therefore, $x$ lies in the interval $\left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)or\left( {\dfrac{\pi }{2},\pi } \right)$
The final answer is $\left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)or\left( {\dfrac{\pi }{2},\pi } \right)$
Hence, option $(1)$ is the correct answer.
So, the correct answer is “Option 1”.

Note: The question involves both trigonometric and inverse trigonometric functions. So, learn the values for each of them. They all have different ranges. So, memorize them as well. The inverse and the trigonometric functions cancel each other out if they are similar. Check which function will cancel out and then multiply it on both sides.
Inverse trigonometric functions produce the length of arc needed to obtain a particular value. The inverse trigonometric functions perform the opposite operations of the trigonometric functions, sine, cosine, tangent, cotangent, secant, cosecant. They are also called arc functions. They have their own domains and ranges for each particular function.