What is $tan (π + arcsin (\frac{2}{3}))$ $tan(pi + arcsin (2/3))$ ?

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What is $tan (π + arcsin (\frac{2}{3}))$ $tan(pi + arcsin (2/3))$ ?

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Answer 1 · 2015-07-21T23:15:24

$\frac{2 \sqrt{5}}{5}$ $(2sqrt(5))/5$

Explanation:

First thing to note is that every $tan$ $color(red)tan$ function has a period of $π$ $pi$

This means that $tan (π + angle) \equiv tan (angle)$ $tan(pi+color(green)"angle")-=tan(color(green)"angle")$

$\Rightarrow tan (π + arcsin (\frac{2}{3})) = tan (arcsin (\frac{2}{3}))$ $=>tan(pi+arcsin(2/3))=tan(arcsin(2/3))$

Now, let $θ = arcsin (\frac{2}{3})$ $theta=arcsin(2/3)$

So, now we are looking for $tan (θ)!$ $color(red)tan(theta)!$

We also have it that : $sin (θ) = \frac{2}{3}$ $sin(theta)=2/3$

Next, we use the identity : $tan (θ) = \frac{sin (θ)}{cos (θ)} = \frac{sin (θ)}{\sqrt{1 - {sin}^{2} (θ)}}$ $tan(theta)=sin(theta)/cos(theta)=sin(theta)/sqrt(1-sin^2(theta))$

And then we substitute the value for $sin (θ)$ $sin(theta)$

$\Rightarrow tan (θ) = \frac{\frac{2}{3}}{\sqrt{1 - {(\frac{2}{3})}^{2}}} = \frac{2}{3} \times \frac{1}{\sqrt{1 - \frac{4}{9}}} = \frac{2}{3} \times \frac{1}{\sqrt{\frac{9 - 4}{9}}} = \frac{2}{3} \times \sqrt{\frac{9}{9 - 4}} = \frac{2}{3} \times \frac{3}{\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$ $=>tan(theta)=(2/3)/sqrt(1-(2/3)^2)=2/3xx1/sqrt(1-4/9)=2/3xx1/sqrt((9-4)/9)=2/3xxsqrt(9/(9-4))=2/3xx3/sqrt(5)=2/sqrt(5)=(2sqrt(5))/5$

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