Posted by **Vincent** on Thursday, June 4, 2009 at 11:23pm.

Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5. (Both u and v are in Quadrant II.)

Find csc(u-v).

First of all, I drew the triangles of u and v. Also, I know the formula of sin(u-v) is sin u * cos v - cos u * sin v. However, I am confused of what you do when you have csc(u-v). Would you do csc u * sec v - sec u * csc v?

Thanks.

- Alg2/Trig -
**Reiny**, Thursday, June 4, 2009 at 11:37pm
no, that is wishful thinking.

csc(u-v) = 1/sin(u-v)

so work out the sin(u-v) according to your formula, the take the reciprocal of your final fraction.

- Alg2/Trig -
**Vincent**, Thursday, June 4, 2009 at 11:42pm
Okay, I worked sin(u-v) out and I got 33/65. Would I then flip it over and get 65/33 as my final answer?

- Alg2/Trig -
**Reiny**, Thursday, June 4, 2009 at 11:44pm
yes, that is also what I got

(and I checked it by finding the actual angles)

- Alg2/Trig -
**Vincent**, Thursday, June 4, 2009 at 11:45pm
Okay thanks for your help!

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