Trig Identity Compound Angle Question where sinA + cosB + sin C = 0 = cos A + sin B + cos C

Question

I'm having trouble proving that sin(A+B) + sin(B+C) + cos(C-A) = -3/2

given that sinA + cosB + sin C = 0 = cos A + sin B + cos C

I've gone around in circles with the trig identities but have been unable to arrive at the correct solution. Any help is appreciated.

See also :https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question — lab bhattacharjee, Feb 03 '20 at 14:43
Bro square both equations and add them , and hence you get your required result. — aryan bansal, Feb 03 '20 at 14:46
Still not getting it. I ended up with -1/2 = -cos^2A +cos^2B - cos^2C
.Excuse my ignorance, I haven't tried this stuff in years. — John, Feb 03 '20 at 18:37

score 0 · Answer 1 · answered Feb 03 '20 at 14:41

0

Hint

We can actually prove something more!

$$(\sin A+\cos B)^2+(\cos A+\sin B)^2=?$$

Similarly $$(\sin A+\sin C)^2+(\cos A-\cos C)^2=?$$

answered Feb 03 '20 at 14:41

lab bhattacharjee

1 Answers1