How do you simplify $sin (2 \cdot arcsin (x))$ $sin (2 * arcsin (x))$ ?

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Answer 1 · 2016-11-08T20:11:19

The answer is $= 2 x \sqrt{1 - x^{2}}$ $=2xsqrt(1-x^2)$

Let $y = arcsin x$ $y=arcsinx$ , then $x = sin y$ $x=siny$

$sin (2 arcsin x) = sin 2 y = 2 sin y cos y$ $sin(2arcsinx)=sin2y=2sinycosy$

${cos}^{2} y + {sin}^{2} y = 1$ $cos^2y+sin^2y=1$

${cos}^{2} y = 1 - x^{2}$ $cos^2y=1-x^2$ $\Rightarrow$ $=>$ $cos y = \sqrt{1 - x^{2}}$ $cosy=sqrt(1-x^2)$

$:.sin(2arcsinx)=2xsqrt(1-x^2)$

Answer 2 · 2018-06-15T23:32:19

If we interpret $arcsin a$ as all the solutions to $sin x = a$ then

$sin(2 arcsin x) = 2 (sin arcsin x)(cos arcsin x) = 2 x cos arcsin (x/1) = pm 2x sqrt{1-x^2}$

$arcsin (x/1)$ refers to a right triangle, opposite $x$ , hypotenuse $1$ so adjacent $sqrt{1-x^2}$ . The sign is ambiguous so we prepend $pm$ .

How do you simplify sin (2 * arcsin (x))sin(2⋅arcsin(x))sin (2 * arcsin (x))?