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What is Sin( inverse COS 3/5)?

1 Answer

Sep 14, 2015

$\frac{4}{5}$ $4/5$

Explanation:

Let inverse $cos (\frac{3}{5})$ $cos(3/5)$ = $θ$ $theta$ , so that it is ${cos}^{- 1} (\frac{3}{5}) = θ$ $cos^-1 (3/5)= theta$

That means $cos θ = \frac{3}{5}$ $cos theta = 3/5$ .

Now work out $sin θ = \sqrt{1 - {cos}^{2} θ} = \frac{4}{5}$ $sin theta = sqrt(1-cos^2 theta)= 4/5$

$θ = {sin}^{- 1} (\frac{4}{5})$ $theta = sin^-1 (4/5)$

${sin cos}^{- 1} (\frac{3}{5}) = {sin sin}^{- 1} (\frac{4}{5}) = \frac{4}{5}$ $sin cos^-1 (3/5)= sin sin^-1 (4/5)= 4/5$

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