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Can you prove the trig identity: sec x = (sin 2x) / (sin x) - (cos 2x) / (cos x) ?
I was able to get an answer of 1 / 2 / cos x, which is so close that it is agonizing! Please help!
Thank you!
3 Answers
- Iggy RockoLv 77 years agoFavorite Answer
sin(2x)/sinx - cos(2x)/cosx =
2sinxcosx/sinx - (cos^2x - sin^2x)/cosx =
2cosx - cos^2x/cosx + sin^2x/cosx =
2cosx - cosx + (1 - cos^2x)/cosx =
cosx + 1/cosx - cos^2x/cosx =
cosx + secx - cosx =
secx
- 7 years ago
sec x = 1/cos x
sin 2x = 2 sin x cos x --------> sin 2x/sin x = 2 cos x
cos 2x = 2 cos^2 x - 1 ---------> cos 2x/cos x = 2 cos x - 1/cos x
sec x = 2 cos x - (2 cos x - 1/cos x) = 1/cos x
- L. E. GantLv 77 years ago
rhs
= (sin(2x)/sin(x)) - cos(2x)/cos(x)
= (2sin(x)cos(x)/sin(x)) - (2cos^2(x) -1)/cos(x)
= 2cos(x) - 2cos(x) + 1/cos(x)
= 1/cos(x)
= sec(x)
= lhs
