# Quick and Easy Question

1. Sep 29, 2007

### Karma

For the question tan (cos-(1/3))... is it possible to represent the value of (1/3) in terms of radians? if not would the answer be tan(70.52877939)

(- being the inverse of cos)

Last edited: Sep 29, 2007
2. Sep 29, 2007

### EnumaElish

a = cos-1(x) or a = arccos(x) is the solution to x = cos(a). The argument of cos (that is a) can be in radians, but x is a real number.

What is the definition of cosine?

How did you get tan(70.52877939)?

3. Sep 29, 2007

### Karma

Ahhh...well after looking at it carefully ..i have derived this as my solution...please tell me if i did it right or wrong..

tan (cos-(1/3))

pheta= @

cos- means the return of an angle. so...cos(x)=@ than cos-(cos@)=0
@=cos- (1/3) --> cos@=1/3

cos@=1/3

cos is the ratio of the lengths of adjacent and hypotneuse of the triangle

to find the other side which is the opposite side.. i used the pythagorean theorm which is
the Hypotneuse^2 - Adjacent^2 and this becomes...

3^2 -1^2 ( all over square root)
so the square root of 8 is the opposite

and to find the tangent angle now is easy since i have the opposite and adjacent side

this becomes tan@= sqrt(8)/1

therefore the answer is tan (cos-(1/3))=tan@=sqrt(8)/1

is my solution correct? and thanks for the help.

4. Sep 30, 2007

### HallsofIvy

Staff Emeritus
If you want to do it more formally, you can use trig identities:
$$tan(\theta)= \frac{sin(\theta)}{cos(\theta)}= \frac{\sqrt{1- cos^2(\theta)}}{cos(\theta)}$$
Since know that $\theta= cos^{-1}(1/3)$, $cos(\theta)= 1/3$ so
$$tan(cos^{-1}(1/3))= \frac{\sqrt{1- \frac{1}{9}}}{\frac{1}{3}}$$
$$= \frac{\frac{\sqrt{8}}{3}}{\frac{1}{3}}= \sqrt{8}= 2\sqrt{2}$$

But using setting it up in terms of a triangle is simpler and better.