Find the exact solutions of the equation cos 2a=square root(1-sin^2 a), 0<a<180.
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We have to determine the solutions for cos 2a = sqrt( 1 - sin a)^2, lying in the range 0<a<180
cos 2a = sqrt( 1 - (sin a)^2)
=> cos 2a = sqrt ( (cos a)^2)
=> cos 2a = cos a
=> 2*(cos a)^2 - 1 = cos a
=> 2*(cos a)^2 - cos a - 1 = 0
=> 2*(cos a)^2 - 2cos a + cos a - 1 = 0
=> 2*cos a ( cos a + 1) + 1(cos a + 1) = 0
=> (2cos a + 1)( cos a + 1) = 0
2*cos a + 1 = 0
=> cos a = -1/2
=> a = arc cos (-1/2)
=> 120 degrees
cos a = 1
=> a = arc cos (1)
=> a = 0, but it does not lie in the interval.
The required value of a is 120 degrees.
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First, we'll change the equation, using the double angle formula, to the left side and the fundamental formula, for the right side:
cos 2a = 2(cos a)^2 - 1
1 - (sin a)^2 = (cos a)^2
2(cos a)^2 - 1 = sqrt[(cos a)^2]
2(cos a)^2 - 1 = cos a
We'll move cos a to the left side:
2(cos a)^2 - cos a - 1 = 0
We'll put cos a = t
2t^2 - t - 1 = 0
We'll apply quadratic formula:
t1 = [1 + sqrt(1 + 8)]/4
t1 = (1+3)/4
t1 = 1
t2 = -1/2
cos a = t1 <=> cos a = 1
a = 0 or a = 180
cos a = -1/2
a = pi - pi/3
a = 2pi/3
SinceĀ 0<a<180, the only solution for the equation is 2*pi/3 = 120 degrees.
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