Asked by jh
Solve:
sin(theta)+cos(theta) = 1
0 < or equal to theta < 2pi
Solution Set:__________ ?
sin(theta)+cos(theta) = 1
0 < or equal to theta < 2pi
Solution Set:__________ ?
Answers
Answered by
Damon
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
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
Answered by
Damon
Oh, and then only when the one that is not zero is +1, not -1
Answered by
jh
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?
Answered by
Damon
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.
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