Posted by Constance on Tuesday, November 6, 2012 at 9:35am.
2-sin^20/cos0 = sec0 + cos0
* 0 = Theta
- Math - Reiny, Tuesday, November 6, 2012 at 10:05am
ok, so it is
2 - sin^2 Ų/cosŲ = secŲ + cosŲ ??
2 - sin^2 Ų/cosŲ = 1/cosŲ + cosŲ
2cosŲ - sin^2 Ų = 1 + cos^2 Ų
2cosŲ = 1 + sin^2 Ų + cos^2 Ų
2cosŲ = 1+1
cosŲ = 1
Ų = 0 , 2π or x = 0° , 360° , for the first period
x = 2kπ, where k is an integer
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