# Thread: Calculating values when combining the tangens and cosinus function

1. ## Calculating values when combining the tangens and cosinus function

Hello, I'm currently studying for an exam I will have right before semester 2 starts now in August and I could really use some help

The question is basically "Calculate the value of tan(cos^-1(0.5))"

I can't understand why or how I would do this, do I turn cosinus into something using sinus and tangens or where do I begin??

Thank you!

2. ## Re: Calculating values when combining the tangens and cosinus function

Hello, I'm currently studying for an exam I will have right before semester 2 starts now in August and I could really use some help

The question is basically "Calculate the value of tan(cos^-1(0.5))"
First, $cos^{-1}(0.5)$. Ask yourself this...The cosine of what angle is 0.5? (Hint: It's one of the "common" angles.)

-Dan

3. ## Re: Calculating values when combining the tangens and cosinus function

Alright, so cos(0.5) = 60deg = 1pi/3

tan(60) = sqrt(3) correct?

and how is this calculated without a calculator, is it possible or do I have to memorize it?

4. ## Re: Calculating values when combining the tangens and cosinus function

Alright, so cos(0.5) = 60deg = 1pi/3

tan(60) = sqrt(3) correct?

and how is this calculated without a calculator, is it possible or do I have to memorize it?
It's a good idea to know the sine, cosine, and tangents of 0, pi/6, pi/4, pi/3, pi/2. (0, 30, 45, 60, 90 in degrees.) They come up all the time in Trig.

One correction: cos(0.5) is not 60. You meant that cos(60) = 0.5 Otherwise, yes, you have it right!

-Dan

5. ## Re: Calculating values when combining the tangens and cosinus function

tan(x)= sin(x)/cos(x) so tan(arcos(x))= sin(arcos(x))/cos(arcos(x)). The denominator is just cos(arcos(x))= x. For the numerator, $sin^2(x)+ cos^2(x)= 1$ so that $sin(x)= \pm\sqrt{1- cos^2(x)}$ and $sin(arcos(x))= \pm\sqrt{1- cos^2(arcos(x)}= \pm\sqrt{1- x^2}$. The sign is the sign on x itself.
A less "rigorous" way to do this is to imagine a right triangle with hypotenuse of length 1 and "near side" of length x (so that the cosine of the angle is x/1= x). By the Pythagorean theorem the "opposite side" has length $\sqrt{1- x^2}$. tangent is "opposite side over near side" so $\sqrt{1- x^2}/x$. Again, if you are allowing angles outside 0 to $\pi/2$, you will need to check the sign.