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

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

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Answer 1 · 2018-03-05T14:59:40

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

Explanation:

Let ${cos}^{- 1} (\frac{1}{2}) = x$ $cos^(-1)(1/2)=x$ then $cos x = \frac{1}{2}$ $cosx=1/2$

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

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

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

Answer 2 · 2018-03-05T15:13:41

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

Explanation:

To find value of $sin ({cos}^{- 1} (\frac{1}{2}))$ $sin (cos ^-1 (1/2))$

Let theta = cos^-1 (1/2)#

$cos θ = (\frac{1}{2})$ $cos theta = (1/2)$

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We know, from the above table, $cos 60 = \frac{1}{2}$ $cos 60 = 1/2$

Hence theta = 60^@#

Replacing ${cos}^{- 1} (\frac{1}{2})$ $cos^-1 (1/2)$ with $θ = 60^{\circ}$ $theta = 60^@$ ,

The sum becomes, $\Rightarrow sin θ = sin 60 = \frac{\sqrt{3}}{2}$ $=> sin theta = sin 60 = sqrt3 / 2$ (As per table above)

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

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