Find the sum of the three smallest positive values of theta such that 4(cos^2)(2theta-pi) =3. (Give your answer in radians.)
LaTEX: Find the sum of the three smallest positive values of $\theta$ such that $4\cos^2(2\theta-\pi) =3$. (Give your answer in radians.)
4cos^2(2θ-π) = 3
cos^2(2θ-π) = 3/4
since cos(a-b) = cosa cosb + sina sinb, cos(2θ-π) = -cos(2θ)
cos^2(2θ) = 3/4
cos(2θ) = ±√3/2
now cos π/6 = √3/2
so, 2θ = π/6, 5π/6, 7π/6, 11π/6
θ = π/12, 5π/12, 7π/12, ...
their sum is 13π/12
To find the three smallest positive values of θ, we first need to solve the equation 4(cos^2)(2θ - π) = 3.
Step 1: Simplify the equation
Use the identity cos^2(x) = 1/2 * [1 + cos(2x)] to simplify the equation:
4 * (1/2 * [1 + cos(2(2θ - π))]) = 3.
Simplifying further, we have:
2 * [1 + cos(4θ - 2π)] = 3.
Step 2: Solve for cos(4θ - 2π)
Rearrange the equation to isolate cos(4θ - 2π):
cos(4θ - 2π) = (3 - 2) / 2 = 1/2.
Step 3: Find the possible values of (4θ - 2π)
Using the inverse function of cosine, we have:
4θ - 2π = ± arccos(1/2).
Since the question asks for the three smallest positive values of θ, we will focus on finding the positive solutions.
Step 4: Solve for θ
Solve the equation 4θ - 2π = arccos(1/2) for θ.
For the principal value, we have:
4θ - 2π = π/3.
Solving for θ, we get:
4θ = π/3 + 2π,
θ = (π/3 + 2π)/4.
Step 5: Finding the three smallest positive values of θ
Since the question asks for the three smallest positive values of θ, we can obtain them by adding integer multiples of 2π/4 to the principal value:
θ1 = (π/3 + 2π)/4 = π/12 + π/2,
θ2 = θ1 + 2π/4 = π/12 + π/2 + π/2 = π/12 + π = 13π/12,
θ3 = θ1 + 4π/4 = π/12 + π/2 + 2π/2 = π/12 + 2π = 25π/12.
Thus, the sum of the three smallest positive values of θ is:
θ1 + θ2 + θ3 = (π/12 + π/2) + (13π/12) + (25π/12) = 39π/12 = 13π/4.
Therefore, the sum of the three smallest positive values of θ is 13π/4 radians.