Question

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

Answer 1

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,
$\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8)$ = $2{\text{sin(si}}{{\text{n}}^{ - 1}}0.8)\cos ({\sin ^{ - 1}}0.8)$
Now, 0.8 can be written as $\dfrac{4}{5}$ in the form of fraction. So,
 $\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8)$ = $2{\text{sin(si}}{{\text{n}}^{ - 1}}\dfrac{4}{5})\cos ({\sin ^{ - 1}}\dfrac{4}{5})$ …… (1)
Now, let ${\sin ^{ - 1}}\dfrac{4}{5}{\text{ }} = {\text{ }}\alpha $, so, $\sin \alpha {\text{ = }}\dfrac{4}{5}$. Now, by using the property ${\sin ^2}\alpha {\text{ + co}}{{\text{s}}^2}\alpha {\text{ = 1}}$, we get
$\cos \alpha {\text{ = }}\dfrac{3}{5}$
Therefore, $\alpha {\text{ = co}}{{\text{s}}^{ - 1}}\dfrac{3}{5}$
$ \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
$\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8)$ = $2{\text{sin(si}}{{\text{n}}^{ - 1}}\dfrac{4}{5})\cos ({\cos ^{ - 1}}\dfrac{3}{5})$
Now, using the property $\sin ({\sin ^{ - 1}}{\text{x) = x}}$ and $\cos ({\cos ^{ - 1}}{\text{x) = x}}$ when x $ \in $ [-1,1]
As, the value of x is smaller than 1 so, by applying the above property, we get
$\sin (2{\text{si}}{{\text{n}}^{ - 1}}0.8)$ = $2\left( {\dfrac{4}{5}} \right)\left( {\dfrac{3}{5}} \right)$ = 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 $\sin ({\sin ^{ - 1}}{\text{x)}}$ 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.