find an algebraic expression equivalent to the given expression. hint: form a right triangle.\n\\( \\sin…

find an algebraic expression equivalent to the given expression. hint: form a right triangle.\n\\( \\sin \\left( 2 \\sin ^ { - 1 } x \\right) \\)\n\\( \\sin \\left( 2 \\sin ^ { - 1 } x \\right) = \\square \\)\n(simplify your answer. type an exact answer, using radicals as needed. do not rationalize the denominator.)

find an algebraic expression equivalent to the given expression. hint: form a right triangle.\n\\( \\sin \\left( 2 \\sin ^ { - 1 } x \\right) \\)\n\\( \\sin \\left( 2 \\sin ^ { - 1 } x \\right) = \\square \\)\n(simplify your answer. type an exact answer, using radicals as needed. do not rationalize the denominator.)

Answer

Explanation:

Step1: Let (\theta=\sin^{- 1}x)

By the definition of inverse - sine function, if (\theta = \sin^{-1}x), then (\sin\theta=x) and (\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]).

Step2: Use the double - angle formula (\sin(2\theta)=2\sin\theta\cos\theta)

Since (\sin\theta = x), we need to find (\cos\theta). Using the identity (\sin^{2}\theta+\cos^{2}\theta = 1), we get (\cos\theta=\sqrt{1 - \sin^{2}\theta}). Substituting (\sin\theta=x) into it, we have (\cos\theta=\sqrt{1 - x^{2}}) (because when (\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), (\cos\theta\geqslant0)).

Step3: Substitute (\sin\theta) and (\cos\theta) into the double - angle formula

(\sin(2\sin^{-1}x)=\sin(2\theta)=2\sin\theta\cos\theta). Substituting (\sin\theta = x) and (\cos\theta=\sqrt{1 - x^{2}}), we get (2x\sqrt{1 - x^{2}}).

Answer:

(2x\sqrt{1 - x^{2}})