How do you evaluate $cos ({sin}^{- 1} ((\frac{\sqrt{3}}{2}))$ $cos(sin^-1((sqrt3/2))$ without a calculator?

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How do you evaluate $cos ({sin}^{- 1} ((\frac{\sqrt{3}}{2}))$ $cos(sin^-1((sqrt3/2))$ without a calculator?

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Answer 1 · 2017-02-04T19:24:56

$cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})) = \frac{1}{2}$ $cos(sin^(-1)(sqrt(3)/2)) = 1/2$

Explanation:

Consider an equilateral triangle with sides of length $2$ $2$ , bisected to form two right angled triangles:

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Remembering that $sin = \frac{opposite}{hypotenuse}$ $sin = "opposite"/"hypotenuse"$ , we can see that:

$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3) = sqrt(3)/2$

Since $\frac{\sqrt{3}}{2} > 0$ $sqrt(3)/2 > 0$ and $\frac{π}{3}$ $pi/3$ is in Q1, we can deduce:

${sin}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}$ $sin^(-1)(sqrt(3)/2) = pi/3$

From the same diagram, remembering $cos = \frac{adjacent}{hypotenuse}$ $cos = "adjacent"/"hypotenuse"$ , we can see that:

$cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3) = 1/2$

So:

$cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})) = cos (\frac{π}{3}) = \frac{1}{2}$ $cos(sin^(-1)(sqrt(3)/2)) = cos(pi/3) = 1/2$

Answer 2 · 2017-02-04T19:30:54

$cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})) = \frac{1}{2}$ $cos(sin^(-1)(sqrt(3)/2)) = 1/2$

Explanation:

Starting from:

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

Subtract ${sin}^{2} θ$ $sin^2 theta$ from both sides to get:

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

Take the square root to find:

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

If $θ = {sin}^{- 1} (\frac{\sqrt{3}}{2})$ $theta = sin^(-1)(sqrt(3)/2)$ then $sin (θ) = \frac{\sqrt{3}}{2}$ $sin(theta) = sqrt(3)/2$ and we find:

$cos θ = \pm \sqrt{1 - {(\frac{\sqrt{3}}{2})}^{2}} = \pm \sqrt{1 - \frac{3}{4}} = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$ $cos theta = +-sqrt(1-(sqrt(3)/2)^2) = +-sqrt(1-3/4) = +-sqrt(1/4) = +-1/2$

Further note that $\frac{\sqrt{3}}{2} > 0$ $sqrt(3)/2 > 0$ , so ${sin}^{- 1} (\frac{\sqrt{3}}{2})$ $sin^(-1)(sqrt(3)/2)$ must be in Q1. Hence $cos θ > 0$ $cos theta > 0$ and we need to choose the positive square root.

So:

$cos ({sin}^{- 1} (\frac{\sqrt{3}}{2})) = \frac{1}{2}$ $cos(sin^(-1)(sqrt(3)/2)) = 1/2$

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