Sum and difference identities

[Louise Johnson]

Question:
Given: sin a = -4/5 and pi<a<3pi/2 and sin B =12/13 and pi/2<B<pi. Find the exact value of:

#1
sin(a-B)
Answer -36/65 I used sinacosB+cosasinB

#2
tan(a+B)
Answer 16/63 I used cosacosB-sinasinB to get cos(a+B)
Then put sin(a+B) over cos(a+B)

#3
cos(2B)

[/b][/color][/size]This one I am stuck on. Hopefully I am on the correct path with the first two drawing the triangles and finding the remaining side. To solve for B in this one do I multiply cosa with sin B and then plug the answer into cos(2B).
Any help with this would be really appreciated.

Thank you
Louise

 

[skeeter]

sin(A) = -4/5, A is in quad III
cos(A) = -3/5

sin(B) = 12/13, B is in quad II
cos(B) = -5/13

#1 ... sin(A-B) = sinAcosB - cosAsinB = (-4/5)(-5/13) - (-3/5)(12/13) = 56/65

#2 ... try this one again

#3 ... cos(2B) = cos(B + B) = cosBcosB - sinBsinB = cos²B - sin²B

 

[Louise Johnson]

Thank you Skeeter, I made a pretty silly error forgetting that cos a and cosb were negative. ( I had even drawn out my quadrant using the CAST acronym!!)

So my new answers

#1 56/65

#2 56/33

#3
-119/169

I used cosBcosB-SinBSinB I hope that is ok? Please let me know other wise. I didn't have any of that in my sum/difference identities information so I have it written down now. ( I can see how it was created though by simplifying and expanding)

 

[skeeter]

#2 ... once again

tan(A+B) = sin(A+B)/cos(A+B) = (sinAcosB+cosAsinB)/(cosAcosB-sinAsinB) =

(20/65 - 36/65)/(15/65 + 48/65) = -16/63

 

[Louise Johnson]

Thank you skeeter I can see where I went wrong with the negative signs. I get pretty excited and rush too much knowing there is help waiting on the other end of the forum.
Thank you for your patience!
Louise
Ps did I get #3 correct?

 

[skeeter]

#3 is correct