Solveeit Logo
Login

Question

Question: How do you calculate \({\sin ^{ - 1}}\sin \left( 2 \right)\)?...

How do you calculate sin⁡−1sin⁡(2){\sin ^{ - 1}}\sin \left( 2 \right)?

Explanation

Solution

In this question we have to find the value of inverse of sine, first assume the given function as a variable, let it be yy, and simplify the expression, and we should know that the range of principal value of sin⁡−1{\sin ^{ - 1}} is equal to [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right], and then using the fact that sin positive in first and second quadrant we will get the required value of the given expression.

Complete step-by-step solution:
Given expression is sin⁡−1sin⁡(2){\sin ^{ - 1}}\sin \left( 2 \right),
Let the expression be equal to yy, i.e.,
y=sin⁡−1sin⁡(2)y = {\sin ^{ - 1}}\sin \left( 2 \right),
Now taking sin on both sides we get,
⇒sin⁡y=sin⁡(sin⁡−1sin⁡(2))\Rightarrow \sin y = \sin \left( {{{\sin }^{ - 1}}\sin \left( 2 \right)} \right),
Now we know that sin⁡(sin⁡−1x)=x\sin \left( {{{\sin }^{ - 1}}x} \right) = x, now applying the fact we get,
⇒sin⁡y=sin⁡(2)\Rightarrow \sin y = \sin \left( 2 \right),
We know that the range value of sin⁡−1{\sin ^{ - 1}} is equal to [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right], but here we got y=2y = 2, which is not in the range of sin value, and we also know that fact that sin positive in first and second quadrant and negative in third and fourth quadrant,
So, sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta ,
Now applying the fact we get,
⇒sin⁡y=sin⁡2\Rightarrow \sin y = \sin 2,
Now 2 can be rewritten as,
⇒sin⁡y=sin⁡(π−2)\Rightarrow \sin y = \sin \left( {\pi - 2} \right),
Now taking sin inverse on both sides we get,
⇒sin⁡−1(sin⁡y)=sin⁡−1(sin⁡(π−2))\Rightarrow {\sin ^{ - 1}}\left( {\sin y} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - 2} \right)} \right),
Now we know that sin⁡(sin⁡−1x)=x\sin \left( {{{\sin }^{ - 1}}x} \right) = x, now applying the fact we get,
⇒y=π−2\Rightarrow y = \pi - 2,
Now substituting the value of yy, we get,
⇒sin⁡−1(sin⁡(2))=π−2\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( 2 \right)} \right) = \pi - 2,
Which is in the range of the principal value of sin⁡−1{\sin ^{ - 1}} which is equal to [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right],

∴\therefore The value of the given function sin⁡−1sin⁡(2){\sin ^{ - 1}}\sin \left( 2 \right) is equal to π−2\pi - 2.

Note: The trigonometry and inverse trigonometry are inverse of each other, and they are represented by arc of the function or the function raised to the power of -1. The sine, cosine, tangent, cosecant, secant, and cotangent are the six types of ratios of trigonometry. The inverse trigonometric functions in trigonometry are used to calculate the angles through any of the trigonometric ratios.