How do you evaluate $cos (arcsin (\frac{3}{5}))$ $cos (arcsin (3/5))$ ?

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How do you evaluate $cos (arcsin (\frac{3}{5}))$ $cos (arcsin (3/5))$ ?

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Answer 1 · 2018-05-26T06:55:54

$cos (arcsin (\frac{3}{5})) = \frac{4}{5}$ $cos(arcsin(3/5)) = 4/5$

Explanation:

The ratio $\frac{3}{5}$ $3/5$ should probably be a clue.

Note that $arcsin (x) \in [- \frac{π}{2}, \frac{π}{2}]$ $arcsin(x) in [-pi/2, pi/2]$ and if $0 < x < 1$ $0 < x < 1$ then $arcsin (x) \in (0, \frac{π}{2})$ $arcsin(x) in (0, pi/2)$ .

So $arcsin (\frac{3}{5})$ $arcsin(3/5)$ is an angle in Q1, which we can consider in a right-angled triangle.

Consider a $3$ $3$ , $4$ $4$ , $5$ $5$ (right) triangle:

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We have:

$sin A = \frac{opposite}{hypotenuse} = \frac{3}{5}$ $sin A = "opposite"/"hypotenuse" = 3/5$

$cos A = \frac{adjacent}{hypotenuse} = \frac{4}{5}$ $cos A = "adjacent"/"hypotenuse" = 4/5$

So:

$cos (arcsin (\frac{3}{5})) = cos (A) = \frac{4}{5}$ $cos(arcsin(3/5)) = cos(A) = 4/5$

Alternatively, we could note more generally that:

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2 theta + sin^2 theta = 1$

Hence:

$cos θ = \pm \sqrt{1 - {sin}^{2} θ}$ $cos theta = +-sqrt(1-sin^2 theta)$

If $θ = arcsin (x)$ $theta = arcsin(x)$ , then $θ \in [- \frac{π}{2}, \frac{π}{2}]$ $theta in [-pi/2, pi/2]$ and hence $cos θ \geq 0$ $cos theta >= 0$ and we find:

$cos (arcsin (x)) = \sqrt{1 - x^{2}}$ $cos(arcsin(x)) = sqrt(1-x^2)$

Answer 2 · 2018-05-27T02:54:14

The inverse sine is multivalued, so

$cos arcsin (\frac{3}{5}) = \pm \sqrt{1 - {(\frac{3}{5})}^{2}} = \pm \frac{4}{5}$ $cos arcsin(3/5) = pm sqrt{1-(3/5)^2} = pm 4/5$

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