Thread: 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.

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.

, and, btw,

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plug and chug ...

Oh, I understand now. Thanks.