Find the value for Cos(theta) if the following conditions hold: Cos(2theta)=1/sqrt2 and 180 degrees < 0 < 270 degrees

Question

This is a trigonometric Identity related question.

Answer 1

The first answer is they way I'd go if you were required to use a double angle identity to solve the exercise.

 

Another way it could be solved is:

 

cos(2θ) = 1/√(2)

 

rationalizing the right side gives

 

cos(2θ) = √(2)/2

From the unit circle, we can find that 2θ must equal the following terms:

 

2θ = 45° + 360°*n (n = 0, 1, 2, ...) for Quadrant I

 

2θ = -45° + 360°*n (n = 0, 1, 2, ...) for Quadrant IV

Solving the above two equations for θ yields the following solution set.

 

θ = 22.5° + 180°*n (n = 0, 1, 2, ...)

 

θ = -22.5° + 180°*n (n = 0, 1, 2, ...)

For the solution in the third quadrant (180° < θ < 270°), let n = 0, 1, etc in the above solution sets.

 

This will provide you the same answer as above.

 

 

Answer 2

Let x=θ

 

cos(2x) = 1 / √2

 

 

Rationalizing denominator, we get

 

cos(2x) = √(2) / 2

 

 

Since we want to find cos(x), we need to use the trig identity for double-angle identity for cosine.

 

cos²(x) - sin²(x) = √(2) / 2

 

 

We want this equation to be in terms of only cos(x).

 

cos²(x) - (1 - cos²(x)) = √(2) / 2

 

2cos²(x) - 1 = √(2) / 2

 

 

Solve for cos(x) from this equation, we get.

 

2cos²(x) = 1 + √(2) / 2

 

2cos²(x) = (2 + √(2)) / 2

 

 

Multiply both sides of the equation by 1/2.

 

cos²(x) = (2 + √(2)) / 4

 

 

Square-root both sides of the equation.

 

cos(x) = √[2 + √(2)] / 2

 

 

Next, take the inverse on both sides of the equation to get your angle, x_ref.  Since the angle is going to be in the third quadrant, you will evaluate

 

cos(180 + x_ref).