2007-01-02 00:49:30 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I think you must be leaving out some information.
You can use identities like cos2a = cos^a - sin^2a to state the second in terms of cosa and sina, but that's about all I can glean from this at all.
2007-01-02 00:55:11 · answer #1 · answered by firefly 6 · 0⤊ 0⤋
sin^2A+sin^2B+sin^2c = 1 - cos^2A +1 - cos^2B + 1 - cos^2C
= 1+1+1 -(cos^2A+cos^B+cos^2C)
= 3- (1)
= 3-1
= 2
cos2A+cos2B+cos2C = 2cos^2A-1 +2cos^2B-1+2cos^2C-1
= 2(cos^2A+cos^2B+cos^2C)- (1+1+1)
=2(1) -(3)
=2 -3
= -1
2007-01-04 16:49:22 · answer #2 · answered by spnchennai 1 · 0⤊ 0⤋
i d k
2007-01-02 00:57:40 · answer #3 · answered by rocky9281 1 · 0⤊ 0⤋
...what?
2007-01-02 00:52:15 · answer #4 · answered by numbuh658 1 · 0⤊ 0⤋