# Find the exact solutions of the equation cos 2a=square root(1-sin^2 a), 0<a<180.

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### 2 Answers

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.**

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.**