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Question: How do you evaluate \({\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)\)?...

How do you evaluate sin⁡−1(sin⁡5π6){\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)?

Explanation

Solution

In this question we have to find the value of the given function which is a trigonometric function, we will make use of inverse trigonometric formulas, first as the given angle will not be range of sin which is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right], so write the angle 5π5\pi as the difference of two angles as 6π−π6\pi - \pi , and the angle becomes π−π6\pi - \dfrac{\pi }{6}, and now using the fact sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta simplify the expression, and then by using the fact sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = xwe will get the required result.

Complete step by step solution:
Given function is sin⁡−1(sin⁡5π6){\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right),
First as the given angle 5π6\dfrac{{5\pi }}{6} is not in the range of sin which is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right], we will write the angle as the difference of two angles, by writing we get,
⇒sin⁡−1(sin⁡5π6)=sin⁡−1(sin⁡6π−π6)\Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{{6\pi - \pi }}{6}} \right),
Now simplifying the expression we get,
⇒sin⁡−1(sin⁡5π6)=sin⁡−1(sin⁡(6π6−π6))\Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{6\pi }}{6} - \dfrac{\pi }{6}} \right)} \right),
Now again simplifying by eliminating the like terms, we get,
⇒sin⁡−1(sin⁡5π6)=sin⁡−1(sin⁡(π−π6))\Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi }{6}} \right)} \right),
Now using the trigonometric equality fact that is sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta , we get,
⇒sin⁡−1(sin⁡5π6)=sin⁡−1(sin⁡π6)\Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right),
Now using the inverse trigonometric formula of sine i.e., sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x,
Here x=π6x = \dfrac{\pi }{6}, by substituting the value we get,
⇒sin⁡−1(sin⁡5π6)=π6\Rightarrow {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) = \dfrac{\pi }{6},
So, the value of the given function is π6\dfrac{\pi }{6}.

∴\therefore The value of the given trigonometric function sin⁡−1(sin⁡5π6){\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right) will be equal to π6\dfrac{\pi }{6}.

Note: The inverse trigonometric functions are used to find the unknown measure of an angle of a right triangle when two side lengths are known. Basic inverse trigonometric functions are sin⁡−1x{\sin ^{ - 1}}x, cos⁡−1x{\cos ^{ - 1}}x and tan⁡−1x{\tan ^{ - 1}}x.Specifically, they are the inverses of the sine, cosine, tangent. Some important formulas related to Inverse trigonometry are given below:
sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}\left( {\sin x} \right) = x,
cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x,
tan⁡−1(tan⁡x)=x{\tan ^{ - 1}}\left( {\tan x} \right) = x,
sec⁡−1(sec⁡x)=x{\sec ^{ - 1}}\left( {\sec x} \right) = x,
csc⁡−1(csc⁡x)=x{\csc ^{ - 1}}\left( {\csc x} \right) = x,
cot⁡−1(cot⁡x)=x{\cot ^{ - 1}}\left( {\cot x} \right) = x,
sin⁡−1x=csc⁡−11x{\sin ^{ - 1}}x = {\csc ^{ - 1}}\dfrac{1}{x},
cos⁡−1x=sec⁡−11x{\cos ^{ - 1}}x = {\sec ^{ - 1}}\dfrac{1}{x},
tan⁡−1x=cot⁡−11x{\tan ^{ - 1}}x = {\cot ^{ - 1}}\dfrac{1}{x},
sin⁡−1x=cos⁡−11−x2=tan⁡−1(x1−x2){\sin ^{ - 1}}x = {\cos ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right),
cos⁡−1x=sin⁡−11−x2=tan⁡−1(1−x2x){\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 - {x^2}} }}{x}} \right),
tan⁡−1x=sin⁡−1(x1+x2)=cos⁡−1(11+x2){\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right).