Question

What is $cos\left( 2arcsin\left( -\dfrac{1}{2} \right) \right)$?

Answer 1

Hint: We have cosine and sine inverse functions here. We resolve the brackets starting from the innermost bracket. We then resolve the brackets further. We will use some formulae involving trigonometric functions. We need to be aware about the range of inverse functions as well. We will be using some properties of the cosine function too.

Complete step-by-step answer:
We are given$cos\left( 2arcsin\left( -\dfrac{1}{2} \right) \right)$. Now, we first solve the innermost bracket. For that we need to be aware about the range of the arcsine function. The arcsine function is basically the inverse sin function that works in the following way:
If $sinx=y$, then $x=arcsiny$ or in other notation, we can also write:
$x=sin^{-1}y$
The range of $arcsin\left(x\right)$ is $\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$.
So, we need to find a value in this range such that
$sin\left(x\right)=\dfrac{-1}{2}$
Since we know that
$sin\left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}$
This happens due to the following result:
$\sin \left( -x \right)=-\sin \left( x \right)$
Using this we obtain:
$x = \dfrac{-\pi}{6}$
So, now our problem becomes:
$cos\left(2\times\dfrac{-\pi}{6}\right)=cos\left(\dfrac{-\pi}{3}\right)$
Now, we use the following result:
$cos\left(-\theta\right)=cos\left(\theta\right)$
And we obtain the following:
$\Rightarrow cos\left( 2\times \dfrac{-\pi }{6} \right)=cos\left( \dfrac{-\pi }{3} \right)=cos\left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$
Hence, we have obtained the result after solving the expression.

Note: You should be aware about the basic formulae of trigonometry as well as inverse trigonometry if you want to solve such problems with as minimal errors as possible. Moreover, you should solve the innermost brackets first. And also note that while resolving the brackets, when the sine bracket will be opened then the negative sign will come out whereas when the cosine bracket will open, a positive sign will come out. Don’t mix them up otherwise it will lead to an invalid solution.