SOLUTION: Solve the equation for solutions over the interval [0, 360). Give solutions to the nearest tenth as appropriate.
Cos^2(theta) = sin^2(theta) + 1
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Question 870615: Solve the equation for solutions over the interval [0, 360). Give solutions to the nearest tenth as appropriate.
Cos^2(theta) = sin^2(theta) + 1
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
First solve for theta
cos^2(theta) = sin^2(theta) + 1
cos^2(theta) - sin^2(theta) = 1
cos(2theta) = 1
2theta = arcos(1)
2theta = 0+360*n
2theta = 360*n
theta = 360*n/2
theta = 180n
So theta = 180n generates all of the solutions for theta where n is an integer.
If n = 0, then
theta = 180n
theta = 180*0
theta = 0
If n = 1, then
theta = 180*n
theta = 180*1
theta = 180
If n = 2, then
theta = 180*n
theta = 180*2
theta = 360 ... toss this solution since it's not in the interval [0, 360)
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The only two solutions are theta = 0, theta = 180 (both in degrees)
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