Need help with this 2 Cos Sin problems

Try substituting sin(y) for x in the second problem.

Law of Cosines is the key

for any triangle

c^2 = a^2 + b^2 -2ab(cos(C))
b^2 = c^2 + c^2 -2ac(cos(B))
a^2 = b^2 + c^2 -2bc(cos(A))

Solving for Cos(A), Cos(B) and Cos(C)

Cos(A) = (-a^2+b^2+c^2)/2bc
Cos(B) = (a^2-b^2+c^2)/2ac
Cos(C) = (a^2+b^2-c^2)/2ab

Plug these in for Cos(A), Cos(B) and Cos(C) in the left side of the equation and simplify, you will get the same thing as the right.

for an explanation of why the law of cosines is true, just search for it on the web.

hehe, darnit fachatta you beat me to it. I just noticed

And I stayed up an extra hour looking for the answer fachatta came up with an hour before I logged back in. Nuts.

Now a follow on question. Just exactly what is the cos C in the problem. That is: Cos A = b/c, Cos B=a/c. But which ratio is Cos C? a/b or b/a?

Kelly

ABC are the angles
abc are the edges opposite their respective angles

Everyone sing along....now I know my ABCs, cosines are no mystery.

now I know my ABCs
cosines are no mystery!
but what about the laws of sine?
I'll teach you that another time!

Nice one stuh.

OK Stuh,

I knew the angles had capital letters and the opposite sides have small letters.

In a right triangle the cosA is b/c, cos B is a/c. Adjacent over hyponenuse.

The question remains. For this problem which is the cosC? b/a or a/b?

Kelly

Cos(A, B, or C) is not a simple ratio of two sides becasue this problem does not specify the triangle is a right triangle.

Cos(C) = (a^2+b^2-c^2)/2ac

Perhaps I must give up hope. I knew your answer, and it had already been posted. But I was hoping for a very closed, very short formula.

Kelly