Question

Trig question

Answer 1

SOHCAHTOA

Answer 2

not helpful when im typing a question which wont use that

Answer 3

Let A, B be less than 180 degrees
show that (sinA+sinB)/2 is less than or equal to sin (A+B)/2

Answer 4

Worth a shot. :P

Answer 5

?

Answer 6

Did openstudy break my LaTeX?

Trying again:

I don't think it is. Cosine is strictly decreasing on \((0,\frac{\pi{}}{2})\) and positive. Sin is strictly increasing on \((0,\frac{\pi{}}{2})\) and positive. Suppose both A and B are from the first quadrant:

\[sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B) \le \sin(A)*1+ 1*\sin(B) = \sin(A)+ \sin(B)\]

Answer 7

The end should read \(= \sin{(A)} + \sin{(B)}\)

Answer 8

y only the first quadrant

Answer 9

Anyway, you probably meant to reverse the polarity of the inequality. In order to prove the reverse, just work out the three cases 1) both A and B from first quadrant. 2) both A and B from second quadrant. and 3) one from first and one from second quadrant.

Answer 10

I was just doing the simple disproof so I took the simplest case.