Finding exact value of trig expression given sides

Hello. I am new to this forum.

The problem is:

Find cos (s-t) given cos s = -1/2, with s in quadrant III, and cos t = -3/5, with t in quadrant III.

I found s = 60 deg. I am having trouble finding t though. In the triangle with angle t, -4 is the "length" of the side opposite t. So the lengths of triangle t are 3, 4, and 5. I found that out through the pythagorean theorem. I think the main thing I don't understand may be how to relate these lengths to the properties of the two special right triangles.

Re: Finding exact value of trig expression given sides

Quote:

Originally Posted by

**VincentPaul** Hello. I am new to this forum.

The problem is:

Find cos (s-t) given cos s = -1/2, with s in quadrant III, and cos t = -3/5, with t in quadrant III.

I found s = 60 deg. I am having trouble finding t though. In the triangle with angle t, -4 is the "length" of the side opposite t. So the lengths of triangle t are 3, 4, and 5. I found that out through the pythagorean theorem. I think the main thing I don't understand may be how to relate these lengths to the properties of the two special right triangles.

$\displaystyle \cos{s} = -\frac{1}{2}$ , $\displaystyle \sin{s} = -\frac{\sqrt{3}}{2}$ and, btw, $\displaystyle s = 240^\circ$

if $\displaystyle \cos{t} = -\frac{3}{5}$ , $\displaystyle \sin{t} = -\frac{4}{5}$

$\displaystyle \cos(s-t) = \cos{s}\cos{t} + \sin{s}\sin{t}$

plug and chug ...

Re: Finding exact value of trig expression given sides

Oh, I understand now. Thanks.