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Jlenok [28]
3 years ago
5

Find the exact value of tan (arcsin (two fifths))

Mathematics
1 answer:
aev [14]3 years ago
5 0
There are 2 ways to do this:

1) Transform tan into terms of sin.
tan x = \frac{sin x}{cos x} = \frac{sin x}{\sqrt{1-sin^2 x}}
where sin x = sin(sin^{-1} (\frac{2}{5})) = \frac{2}{5}
Substituting back in gives:
tan x = \frac{\frac{2}{5}}{\sqrt{1-(\frac{2}{5})^2}} = \frac{\frac{2}{5}}{\sqrt{\frac{21}{25}}} = \frac{2}{5}*\frac{\sqrt{25}}{\sqrt{21}} = \frac{2}{\sqrt{21}}


2) Use a right triangle.
\theta = sin^{-1} (\frac{2}{5}) \\  \\ sin \theta = \frac{2}{5}
sin = opp/hyp --> opp = 2, hyp = 5
Use Pythagorean theorem to solve for adjacent side.
adj = \sqrt{5^2 - 2^2} = \sqrt{21}
tan = opp/adj
tan \theta = \frac{2}{\sqrt{21}}
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we are given

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