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
Solve:
sin(theta)+cos(theta) = 1
0 < or equal to theta < 2pi
Solution Set:__________ ?
4 answers
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?
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.
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.