Help please?! Given: sinA=-3/5, cosB=15/17, angle A in Quadrant III and angle B in Quadrant I.
cos2B =
a. -64/289
b. 161/289
c. 225/289
d. 240/289

Question

Help please?! Given: sinA=-3/5, cosB=15/17, angle A in Quadrant III and angle B in Quadrant I.
cos2B = 
a. -64/289
b. 161/289
c. 225/289
d. 240/289

anonymous · Answer

is there a picture for this one?

anonymous · Answer

Just a second.

anonymous · Answer

attachment

anonymous · Answer

I just dont get why they have two different angles here, A and B.  You can just use the same method we did in the last one to find cos(2B).  Look for the identity on the sheet i gave the link for and substitute for it in a similar fashion... sinA just seems to be useless...

**looking at picture

anonymous · Answer

Um, ya.. i don't see how sinA is even needed.  So i'm missing something here or it's just useless info.

anonymous · Answer

look under the double angle formulas for the identity that you need.

anonymous · Answer

i'm sorry but i need to head to bed.  Best of luck!