I'm having trouble understanding this:

tan[arc cos (square root of 3 / 2) = square root of 3 / 3

How does this answer come to be?

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- Sep 20th 2009, 12:49 PMhorsepower.850Inverse Trig Functions
I'm having trouble understanding this:

tan[arc cos (square root of 3 / 2) = square root of 3 / 3

How does this answer come to be? - Sep 20th 2009, 01:17 PMartvandalay11
This question is saying what is the Tangent of the angle who's cosine is

So what is the angle who's cosine equals that. Then take the tangent of your answer....

But even more cleverly,

When cosine equals , sin =

So now compute sin over cos - Sep 20th 2009, 10:08 PMmacosxnerd101RE: Inverse Trig Functions
Let's look at the unit circle: File:Unit circle angles.svg - Wikipedia, the free encyclopedia

As artvandalay said, the first thing the question is asking for is the angle whose cosine is sqrt(3)/2. Knowing that the unit circle coordinates are (cos x, sin x), we can determine that the angle measure whose cosine is sqrt(3)/2 is pi/3. Now from here, the question is asking for the tangent of this angle, which is the sine value over the cosine value. The sine value here is 1/2, and we know the cosine value is sqrt(3)/2. So:

tan pi/3 = (1/2)/(sqrt(3)/2)

The twos cancel out, leaving 1/sqrt(3). When we rationalize the function, we get sqrt(3)/3 (multiply the numerator and denominator by sqrt(3) to remove the radical).

Hope this helps some.