Trigonometry pre calc 12

Question

Solve the equation 2 cos² θ - 1 = 0 for θ, where 0 ≤ θ < 360.

Answer 1

2cos²θ - 1 = 0

Add 1 to both sides:

2cos²θ = 1

Divide both sides by 2:

cos²θ = 1/2

Take the square root of both sides of the equation. Remember that a square root can be positive or negative:

cos(θ) = + sqrt(2)/2

cos(θ) = - sqrt(2)/2

What values of θ make those equations true in the domain 0≤θ≤360?

When θ= 45 or 315, cosθ= sqrt(2)/2.

when θ= 135 or 225, cosθ= - sqrt(2)/2.

Another method is to graph the function in Y1 on the calculator and look for the zeros.

Answer 2

2cos^2x = 1 = cos^2 + sin^2

cos^2x = sin^2x

cosx= sinx or -sinx

x = 45 degrees or pi/4 or 3pi/4 or 5pi/4 or 7pi/4

cos45 = sqr2/2

cos^245 = 2/4 = 1/2

2cos^245 = 2(1/2) = 1