Question

if cscA= -5/3, and cscB= -13/5, what is sin(A+B) if both A/B are in quad 3?

I got -64/65, but that is the wrong answer.. ):

Answer 1

csc A = -5/3 means sinA= -3/5
csc B= -13/5 means sinB= - 5/13
it helps to know (but you can use Pythagoras) that 3,4,5 and 5,12,13 are right triangles

sin(A+B)= sinA cosB + sinB cosA

you know sinA and sinB

cosA is -4/5
cosB is -12/13
(sketch a picture for these)

if we use those values we get
sin(A+B)= -3/5 * -12/13 + -5/13 * -4/5
or
36/65 + 20/65

Answer 2

for sin(A+B), won't it technically become sin(A-B) since -5/13 is negative? I don't really understand how the pythagorean identity comes into play here? thank you for your detailed explanation though

Answer 3

and also, i thought that cos A and B would be 3/5 and 5/13 respectively?

Answer 4

oh, than wouldn't cos be -3/5 and -5/13? I just don't understand why phi uses the pythagorean triangles and why cos aren't those answers instead? since the thing was sin(A+B) which is sin(3/5+5/13)?

Answer 5

yeah

Answer 6

***sin(A+B) which is sin(3/5+5/13)?***
notice that A is not 3/5. sinA is 3/5

Answer 7

let phi continues his stuff, hihihi
@phi please

Answer 8

ohmy @Loser66 that diagram is beautiful! i get the pythagorean thing now, but can @phi continue to explain your thinking? I just need a little clarification :)