Question

Prove that ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) = 2{\cos ^{ - 1}}x,\dfrac{1}{{\sqrt 2 }} \leqslant x \leqslant 1$ .

Answer 1

Hint: By using the basic trigonometric identity given below we can simplify the above expression that is ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) = 2{\cos ^{ - 1}}x,\dfrac{1}{{\sqrt 2 }} \leqslant x \leqslant 1$ . In order to solve and simplify the given expression we have to use the identities and express our given expression in the simplest form and thereby solve it.

Complete step by step solution:
To prove: ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) = 2{\cos ^{ - 1}}x,\dfrac{1}{{\sqrt 2 }} \leqslant x \leqslant 1$
Proof: Taking left hand side of the given equation , we will get the following
 ${\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$
Now we have to simplify the given equation as shown below ,
We have let $x = \cos \theta $ for simplification, substitute this in the above expression , we will get the following result,
$
   \Rightarrow {\sin ^{ - 1}}(2\cos \theta \sqrt {1 - {{\cos }^2}\theta } ) \\
   = {\sin ^{ - 1}}(2\cos \theta \sin \theta ) \\
 $ ( using identity ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ )
Now we have to simplify the given equation as shown below ,
Using the identity $\sin 2\theta = 2\sin \theta \cos \theta $ , we will get ,
$
   = {\sin ^{ - 1}}(\sin 2\theta ) \\
   = 2\theta \\
 $
Now we have to simplify the given equation as shown below ,
$ = 2{\cos ^{ - 1}}x$
Which is equal to R.H.S ,
Since L.H.S=R.H.S
Hence proved.

Note: Some other equations needed for solving these types of problem are:
$1 - {\cos ^2}x = {\sin ^2}x$ and
$\sin 2\theta = 2\sin \theta \cos \theta $
Range of cosine and sine: $\left[ { - 1,1} \right]$ ,
Also, while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.