How do you find the value of $tan [(1/2) cos^-1 (2/3)]$ ?

Question

16712 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-04T19:44:37

$tan(arccos(2/3)/2) = sqrt(5)/5$

From the pythagorean identity we have that

$sin^2x + cos^2x =1$

Dividing both sides by $cos^2x$ we have

$tan^2x + 1 = 1/cos^2x$

Which means that, if we isolate the tangent we have

$tan^2x = 1/cos^2x - 1$

So for $x = arccos(2/3)/2$ we have

$tan^2(arccos(2/3)/2) = 1/cos^2(arccos(2/3)/2) -1$

$cos^2(arccos(2/3)/2)$ can be calculated using the half angle formula, so we know that

$cos^2(arccos(2/3)/2) = (1+cos(arccos(2/3)))/2$

Since $cos(arccos(x)) = x$ we can say that

$cos^2(arccos(2/3)/2) = (1+2/3)/2 = (5/3)/2 = 5/3*1/2 = 5/6$

Putting that back on the formula

$tan^2(arccos(2/3)/2) = 1/(5/6) -1 = 6/5 - 1 = (6-5)/5 = 1/5$

Taking the root

$tan(arccos(2/3)/2) = +-sqrt(1/5)$

Knowing that during the range of the arccosine, the tangent is only negative if the cosine's also negative we have that

$tan(arccos(2/3)/2) = 1/sqrt(5) = sqrt(5)/5$

How do you find the value of tan [(1/2) cos^-1 (2/3)]tan [(1/2) cos^-1 (2/3)]?