Let f:(-1, 1)→R be such that f(cos 4θ) = 2 - sec²θ for θ∈(0, π/4)∪(π/4, π/2). Then the value(s) of f(1/3) is (are) 1 - √3/2, 1 + √3/2, 1

For θ∈(0, π/4)∪(π/4, π/2)
Suppose cos 4θ = 1/3 ⇒ cos 2θ = ±√(1 + cos 4θ)/2 = ±√(2/3)
and, f(1/3) = 2 - sec²θ = 2cos²θ/(2cos²θ -1)
Since cos 2θ = 2cos²θ - 1, then 2cos²θ = 1 + cos 2θ
Therefore, f(1/3) = (1 + cos 2θ)/cos 2θ = 1 + 1/cos 2θ
Then, f(1/3) = 1 ± √3/2

Eduprobe

Question:

Let f:(-1, 1)→R be such that f(cos 4θ) = 2 - sec²θ for θ∈(0, π/4)∪(π/4, π/2). Then the value(s) of f(1/3) is (are) 1 - √3/2, 1 + √3/2, 1 - √2/3, 1 + √2/3

Solution: