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Question: Simplify it. \(\sin \left( {2{{\sin }^{ - 1}}x} \right) = \)?...

Simplify it.
sin⁡(2sin⁡−1x)=\sin \left( {2{{\sin }^{ - 1}}x} \right) = ?

Explanation

Solution

In the above question, first we have to use an identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta by considering θ=sin⁡−1x\theta = {\sin ^{ - 1}}x. Then using the concept of inverse trigonometric function, we will find the value of sin⁡θ\sin \theta and then for cos⁡θ\cos \theta we will use the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 so that we can find the value of cos⁡θ\cos \theta in terms of sin⁡θ\sin \theta whose value is known to us. In the end, we just have to substitute the values.

Complete step-by-step answer:
In the above question, we have to fins the value of sin⁡(2sin⁡−1x)\sin \left( {2{{\sin }^{ - 1}}x} \right).
Here we have to use the formula sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta .
Therefore, on comparing we get θ=sin⁡−1x\theta = {\sin ^{ - 1}}x.
Therefore, we can write the above equation as
sin⁡(2sin⁡−1x)=2sin⁡(sin⁡−1x)cos⁡(sin⁡−1x)\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2\sin \left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right)
Now, we can write as sin⁡(sin⁡−1x)=x\sin ({\sin ^{ - 1}}x) = x when θ∈(π2,−π2)\theta \in \left( {\frac{\pi }{2}, - \frac{\pi }{2}} \right).
Therefore, sin⁡(sin⁡−1x)=x..............(1)\sin ({\sin ^{ - 1}}x) = x..............\left( 1 \right)
Also, we know that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Therefore, cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt {1 - {{\sin }^2}\theta }
We can also write cos⁡(sin⁡−1x)=cos⁡θ\cos \left( {{{\sin }^{ - 1}}x} \right) = \cos \theta and θ=sin⁡−1x\theta = {\sin ^{ - 1}}x
Therefore, cos⁡(sin⁡−1x)=1−x2...............(2)\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} ...............\left( 2 \right)
We have,
sin⁡(2sin⁡−1x)=2sin⁡(sin⁡−1x)cos⁡(sin⁡−1x)\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2\sin \left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right)
Now substitute equation (1) and (2)\left( 1 \right)\,and\,\left( 2 \right) in above equation,
We get,
sin⁡(2sin⁡−1x)=2x1−x2\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2x\sqrt {1 - {x^2}}
Therefore, the required value is 2x1−x22x\sqrt {1 - {x^2}} .

Note: Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trigonometric functions are especially applicable to the right angle triangle.