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Question: Find the value of \(\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8)\) is equal to A. \({\sin ^{ - 1}}1.2...

Find the value of sin⁡(2sin−10.8)\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8) is equal to
A. sin⁡−11.2{\sin ^{ - 1}}1.2
B. sin⁡−1(0.96){\sin ^{ - 1}}(0.96)
C. sin⁡−1(0.48){\sin ^{ - 1}}(0.48)
D. sin⁡1.60\sin {1.6^0}

Explanation

Solution

Hint: Let us use the formula of sin2x and also the property of inverse trigonometric functions.

Complete step-by-step answer:
Now, in this question, first use the property of sin2x= 2 sin x cos x.
Therefore,
sin⁡(2sin−10.8)\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8) = 2sin(sin−10.8)cos⁡(sin⁡−10.8)2{\text{sin(si}}{{\text{n}}^{ - 1}}0.8)\cos ({\sin ^{ - 1}}0.8)
Now, 0.8 can be written as 45\dfrac{4}{5} in the form of fraction. So,
sin⁡(2sin−10.8)\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8) = 2sin(sin−145)cos⁡(sin⁡−145)2{\text{sin(si}}{{\text{n}}^{ - 1}}\dfrac{4}{5})\cos ({\sin ^{ - 1}}\dfrac{4}{5}) …… (1)
Now, let sin⁡−145 = α{\sin ^{ - 1}}\dfrac{4}{5}{\text{ }} = {\text{ }}\alpha , so, sin⁡α = 45\sin \alpha {\text{ = }}\dfrac{4}{5}. Now, by using the property sin⁡2α + cos2α = 1{\sin ^2}\alpha {\text{ + co}}{{\text{s}}^2}\alpha {\text{ = 1}}, we get
cos⁡α = 35\cos \alpha {\text{ = }}\dfrac{3}{5}
Therefore, α = cos−135\alpha {\text{ = co}}{{\text{s}}^{ - 1}}\dfrac{3}{5}
⇒ sin−145 = cos−135\Rightarrow {\text{ si}}{{\text{n}}^{ - 1}}\dfrac{4}{5}{\text{ = co}}{{\text{s}}^{ - 1}}\dfrac{3}{5}
Now, putting this value in equation (1), we get
sin⁡(2sin−10.8)\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8) = 2sin(sin−145)cos⁡(cos⁡−135)2{\text{sin(si}}{{\text{n}}^{ - 1}}\dfrac{4}{5})\cos ({\cos ^{ - 1}}\dfrac{3}{5})
Now, using the property sin⁡(sin⁡−1x) = x\sin ({\sin ^{ - 1}}{\text{x) = x}} and cos⁡(cos⁡−1x) = x\cos ({\cos ^{ - 1}}{\text{x) = x}} when x ∈\in [-1,1]
As, the value of x is smaller than 1 so, by applying the above property, we get
sin⁡(2sin−10.8)\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8) = 2(45)(35)2\left( {\dfrac{4}{5}} \right)\left( {\dfrac{3}{5}} \right) = 0.96
So, the answer is 0.96.

Note: To solve such types of questions we have to use the property of trigonometry and inverse trigonometric function, apply proper property. The property of sin⁡(sin⁡−1x)\sin ({\sin ^{ - 1}}{\text{x)}} gives different results depending on the value of x. If the value of lie in the interval [-1,1] it gives the x, in other cases there are different values where the value of x is in the other interval. In the given question as the value of x is smaller lies in [-1,1], so, it gives x.