Great Trigonometric Equation

Geometry Level 4

Let θ , ϕ ∈ R \theta, \phi \in \mathbb{R} . If cos ⁡ θ − sin ⁡ θ + 1 = sin ⁡ 2 ϕ \sqrt{\cos \theta - \sin \theta} + 1 = \sin 2\phi . Find 2 cos ⁡ 2 θ + 4 sin ⁡ 2 ϕ 2\cos^2 \theta + 4\sin^2 \phi .

3 4 None of these choices 2 1

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Since ϕ ∈ R \phi \in \mathbb{R} we know that sin ⁡ ( 2 ϕ ) ∈ R , \sin(2\phi) \in \mathbb{R}, and thus we will require that ( cos ⁡ ( θ ) − sin ⁡ ( θ ) ) ≥ 0 , (\cos(\theta) - \sin(\theta)) \ge 0, for otherwise the LHS of the equation would have an imaginary component.

But since sin ⁡ ( 2 ϕ ) ≤ 1 \sin(2\phi) \le 1 for any real ϕ \phi we must have that cos ⁡ ( θ ) − sin ⁡ ( θ ) = 0 , \cos(\theta) - \sin(\theta) = 0, for otherwise the LHS of the equation would exceed 1. 1. Thus θ = π 4 + n π \theta = \dfrac{\pi}{4} + n\pi for any integer n , n, and 2 ϕ = π 2 + m ∗ 2 π ⟹ ϕ = π 4 + m π 2\phi = \dfrac{\pi}{2} + m*2\pi \Longrightarrow \phi = \dfrac{\pi}{4} + m\pi for any integer m . m.

We thus find that 2 cos ⁡ 2 ( θ ) + 4 sin ⁡ 2 ( ϕ ) = 2 ∗ 1 2 + 4 ∗ 1 2 = 3 . 2\cos^{2}(\theta) + 4\sin^{2}(\phi) = 2*\dfrac{1}{2} + 4*\dfrac{1}{2} = \boxed{3}.

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