How do you evaluate $tan [arccos (\frac{1}{3})]$ $tan[arccos(1/3)]$ ?

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Answer 1 · 2016-05-03T12:06:46

$tan [arccos (\frac{1}{3})] = 2 \sqrt{2}$ $tan[arccos(1/3)]=2sqrt(2)$

arccos is the reversing of the process of cos to give the angle

$\Rightarrow θ = [arccos (\frac{1}{3}) = arccos (\frac{adjacent}{hypotenuse})]$ $=> theta=[ arccos(1/3) = arccos(("adjacent")/("hypotenuse"))]$

So this is giving us 2 length of sides for a right triangle. From which we can work out the tangent value.
Tony B

By Pythagoras and using the notation in the diagram.

$c^{2} = b^{2} + a^{2} \Rightarrow 3^{2} = 1^{2} + a^{2}$ $c^2=b^2+a^2" " =>" " 3^2=1^2+a^2$

Thus $a = \sqrt{8} = \sqrt{2 \times 2^{2}} = 2 \sqrt{2}$ $a=sqrt(8) = sqrt(2xx2^2)=2sqrt(2)$

$tan (θ) = \frac{opposite}{adjacent} = \frac{a}{b} = \frac{2 \sqrt{2}}{1}$ $tan(theta) = ("opposite")/("adjacent")=a/b = (2sqrt(2))/1$

$tan (θ) = 2 \sqrt{2}$ $tan(theta)=2sqrt(2)$

Thus: $tan [arccos (\frac{1}{3})] = 2 \sqrt{2}$ $" "tan[arccos(1/3)]=2sqrt(2)$

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