Solve:
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.
To solve the equation sin(theta) + cos(theta) = 1, we will use the concept of trigonometric identities and algebraic manipulation.
Step 1: Rearrange the equation.
We can rewrite the given equation as:
sin(theta) = 1 - cos(theta)
Step 2: Substitute using the Pythagorean Identity.
The Pythagorean Identity states: sin^2(theta) + cos^2(theta) = 1.
Substitute sin^2(theta) with (1 - cos^2(theta)):
(1 - cos^2(theta)) + cos(theta) = 1
Step 3: Simplify.
Rearrange and combine like terms:
cos^2(theta) - cos(theta) = 0
Step 4: Factor out common terms.
Factor the equation:
cos(theta)(cos(theta) - 1) = 0
Now, we have two possible cases to consider:
Case 1: cos(theta) = 0
If cos(theta) = 0, then theta must be equal to pi/2 or 3pi/2, since those are the angles where cos(theta) equals zero in the given range of 0 <= theta < 2pi.
Case 2: cos(theta) - 1 = 0
Solve for theta in this case:
cos(theta) = 1
This equation is satisfied when theta = 0, since cos(0) = 1.
Therefore, the solution set for the given equation is:
theta = {0, pi/2, 3pi/2} in the range 0 <= theta < 2pi