Solving the law of cosines for cos(C)

[Paindealer]

Ok I did this thing on law of cosines and here is what i came up with, but i am pretty sure it is wrong so help me change it to solve for cos(C)

c^2=a^2+b^2-2abcos(C)

so...

Cos(C)=____c^2____
. a^2+b^2-2ab

is that right?

if not how do i make it to equal cos(C)?
(need this by like right now cause i am a procrastinator! :P)

 

[mathmike]

wouldnt cos(x) = (a^2 + b^2 - c^2) / 2ab

 

[Paindealer]

probably... i will use that for my homework then, thanks

 

[HallsofIvy]

No! You both are apparently thinking of

c²= (a²+ b²- 2ab) cos(C)
which it is NOT! (Notice the parentheses on the right side.)
The cosine law says c²= a²+ b²- 2ab cos(C). (Notice lack of parentheses!)

The problem is not the "cosine", it is simple algebra:

Suppose the problem were c= a+ bx. How would you solve for x?

 

[ksinclair13]

I don't see how mathmike's is wrong, but I may just be stupid.

Isn't

$$cos (C) = \frac{a^2 + b^2 - c^2}{2ab}$$

the same as

$$cos (C) = \frac{c^2 - a^2 - b^2}{-2ab}$$

?

You could take a -1 out of the numerator and then cancel out the negatives, right?