How do you calculate $sin (2 {sin}^{- 1} (10 x))$ $sin(2sin^-1(10x))$ ?

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Answer 1 · 2015-07-03T18:41:25

$sin (2 {sin}^{- 1} (10 x)) = 20 x \sqrt{1 - 100 x^{2}}$ $sin(2sin^(-1)(10x))=20xsqrt(1-100x^2)$

$Let y = sin (2 {sin}^{- 1} (10 x))$ $"Let " y=sin(2sin^(-1)(10x))$

Now, let $θ = {sin}^{- 1} (10 x) \Rightarrow sin (θ) = 10 x$ $" "theta=sin^(-1)(10x)" "=>sin(theta)=10x$

$\Rightarrow y = sin (2 θ) = 2 sin θ cos θ$ $=>y=sin(2theta)=2sinthetacostheta$

Recall that: ${cos}^{2} θ = 1 - {sin}^{2} θ \Rightarrow cos θ = \sqrt{1 - {sin}^{2} θ}$ $" "cos^2theta=1-sin^2theta=>costheta=sqrt(1-sin^2theta)$

$\Rightarrow y = 2 sin θ \sqrt{1 - {sin}^{2} θ}$ $=>y=2sinthetasqrt(1-sin^2theta)$

$\Rightarrow y = 2 \cdot (10 x) \sqrt{1 - {(10 x)}^{2}} = 20 x \sqrt{1 - 100 x^{2}}$ $=>y=2*(10x)sqrt(1-(10x)^2)=color(blue)(20xsqrt(1-100x^2))$

How do you calculate sin(2sin−1(10x))sin(2sin^-1(10x))?