How do you find the value of tan[(12)cos−1(23)]?

1 Answer
Lovecraft
Oct 4, 2015

tan⎛⎜⎝arccos(23)2⎞⎟⎠=√55

Explanation:

From the pythagorean identity we have that

sin2x+cos2x=1

Dividing both sides by cos2x we have

tan2x+1=1cos2x

Which means that, if we isolate the tangent we have

tan2x=1cos2x−1

So for x=arccos(23)2 we have

tan2⎛⎜⎝arccos(23)2⎞⎟⎠=1cos2(arccos(23)2)−1

cos2⎛⎜⎝arccos(23)2⎞⎟⎠ can be calculated using the half angle formula, so we know that

cos2⎛⎜⎝arccos(23)2⎞⎟⎠=1+cos(arccos(23))2

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

cos2⎛⎜⎝arccos(23)2⎞⎟⎠=1+232=532=53⋅12=56

Putting that back on the formula

tan2⎛⎜⎝arccos(23)2⎞⎟⎠=156−1=65−1=6−55=15

Taking the root

tan⎛⎜⎝arccos(23)2⎞⎟⎠=±√15

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

tan⎛⎜⎝arccos(23)2⎞⎟⎠=1√5=√55

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