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Question: How do you evaluate \(\operatorname{arc}\left( sin\left( \cos \dfrac{\pi }{3} \right) \right)\) ?...

How do you evaluate arc⁡(sin(cos⁡π3))\operatorname{arc}\left( sin\left( \cos \dfrac{\pi }{3} \right) \right) ?

Explanation

Solution

In this question, we have to simplify a trigonometric function. So, we will apply the trigonometric formula to get the solution for the same. Since the question consists of an inverse function, we will first solve the brackets of the inverse function by applying the trigonometric formula cos⁡(π−θ)=−cos⁡θ\cos \left( \pi -\theta \right)=-\cos \theta in the equation. Then, on further solving, we again apply the trigonometric formula cos⁡(π2−θ)=sin⁡θ\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta . In the last, we will again apply the trigonometric formula sin⁡−1(sin⁡θ)=θ{{\sin }^{-1}}\left( \sin \theta \right)=\theta in the equation, to get the required value of the problem.

Complete step by step answer:
According to the question, we have to simplify the trigonometric function
So, we will use the trigonometric formula to get the solution of the problem.
The trigonometric function given to us is arc⁡(sin(cos⁡π3))\operatorname{arc}\left( sin\left( \cos \dfrac{\pi }{3} \right) \right) ----------- (1)
So, first, we will again apply the trigonometric formula cos⁡(π2−θ)=sin⁡θ\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta in the above equation, we get
⇒sin⁡−1(cos⁡(π2−π6))\Rightarrow {{\sin }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{6} \right) \right)
Therefore, we get
⇒sin⁡−1(sin⁡(π6))\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right) \right)
Again we will apply the inverse trigonometric formula sin⁡−1(sin⁡θ)=θ{{\sin }^{-1}}\left( \sin \theta \right)=\theta in the above equation, we get
⇒π6\Rightarrow \dfrac{\pi }{6} which is our required solution.

Therefore, for the equation arc⁡(sin(cos⁡π3))\operatorname{arc}\left( sin\left( \cos \dfrac{\pi }{3} \right) \right) , its simplified value is π6\dfrac{\pi }{6} .

Note: While solving this problem, do mention all the trigonometric formulas to avoid confusion and mathematical errors. One of the alternative methods to solve this problem is putting the cos value from the trigonometric table in the equation and then apply the inverse formula to get the solution of the inverse function.
An alternative method:
The trigonometric function is given to us arc⁡(sin(cos⁡π3))\operatorname{arc}\left( sin\left( \cos \dfrac{\pi }{3} \right) \right)
We will first apply the value of cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2} in the above equation, we get
⇒arcsin⁡(12)\Rightarrow \arcsin \left( \dfrac{1}{2} \right)
⇒sin−1(12)\Rightarrow si{{n}^{-1}}\left( \dfrac{1}{2} \right)
Now, we know that the value of 12\dfrac{1}{2} in the sin function comes at π6\dfrac{\pi }{6} , therefore we get
⇒π6\Rightarrow \dfrac{\pi }{6} which is our required solution.