How do you simplify $sin ({tan}^{- 1} (\frac{1}{2}))$ $sin(tan^-1(1/2))$ ?

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How do you simplify $sin ({tan}^{- 1} (\frac{1}{2}))$ $sin(tan^-1(1/2))$ ?

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Answer 1 · 2018-03-04T07:48:07

We can write as follows,

$sin ({tan}^{- 1} (\frac{1}{2})) = sin ({sin}^{- 1} (\frac{1}{\sqrt{5}})) = \frac{1}{\sqrt{5}}$ $sin(tan^-1(1/2)) =sin(sin^-1(1/(sqrt(5))))=1/sqrt(5)$

Answer 2 · 2018-03-04T07:48:18

The answer is $= \frac{1}{\sqrt{5}}$ $=1/sqrt5$

Explanation:

Let $θ = {tan}^{- 1} (\frac{1}{2})$ $theta=tan^-1(1/2)$

Then,

$tan θ = \frac{1}{2}$ $tantheta=1/2$

We need,

$sin ({tan}^{- 1} (\frac{1}{2})) = sin θ$ $sin(tan^-1(1/2))=sintheta$

If $tan θ = \frac{1}{2}$ $tan theta=1/2$ ,

${tan}^{2} θ + 1 = \frac{1}{{cos}^{2} θ}$ $tan^2theta+1=1/cos^2theta$

${cos}^{2} θ = \frac{4}{5}$ $cos^2theta=4/5$

${sin}^{2} θ = 1 - \frac{4}{5} = \frac{1}{5}$ $sin^2theta=1-4/5=1/5$

$sin θ = \frac{1}{\sqrt{5}}$ $sintheta=1/sqrt5$

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