Precalculus(NEED HELP ASAP PLEASE!!)
posted by jh on .
sin(theta)+cos(theta) = 1
0 < or equal to theta < 2pi
Solution Set:__________ ?
sin T +sqrt (1-sin^2T) = 1
sqrt(1-sin^2T) = 1 - sin T
1 - sin^2 T = 1 - 2 sin t + sin^2 T
-sin^2 T = -2 sin T + sin^2 T
2 sin^2 T = 2 sin T
sin T = 1
T = 0
That tells you something. The only places this happens are when sin = 0 or cos = 0
That is at 0, pi/2, pi, and 3 pi/2
Oh, and then only when the one that is not zero is +1, not -1
Damon I don't understand you 2nd post but is the answer: 0, pi/2, pi, and 3 pi/2 ??
If it is.. then unfortunately its not the correct answer. Do you think you went wrong somewhere?
sin T = +1 when T = 90 degrees , pi/2
cos T = 1 when T = 0 or 2pi
if T = 0 for example
sin T = 0
cos T = 1
and sin T + cos T = 1
So that just plain checks.