Let θ, φ ∈ [0, 2π] be such that 2cosθ (1 - sinφ) = sin²θ (tanθ/2 + cotθ/2)cosφ - 1, tan(2π - θ) > 0 and -1 < sinθ < -√3/2, then φ cannot satisfy

Let θ, φ ∈ [0, 2π] be such that 2cosθ (1 - sinφ) = sin²θ (tanθ/2 + cotθ/2)cosφ - 1, tan(2π - θ) > 0 and -1 0 -1 0 ⇒ tanθ < 0 θ ∈ (3π/2, 5π/3) tanθ/2 + cotθ/2 = 2/sinθ 2cosθ(1-sinφ) = 2sinθcosφ - 1 2cosθ + 1 = 2sinθcosφ + 2cosθsinφ 2cosθ + 1 = 2sin(θ + φ) sin(θ + φ) = cosθ + 1/2 θ ∈ (3π/2, 5π/3) ⇒ cosθ ∈ (0, 1/2) sin(θ + φ) ∈ (1/2, 1) θ + φ ∈ (3π/2, 11π/3) sin s ∈ (1/2, 1) ⇒ s ∈ (13π/6, 17π/6) θ + φ ∈ (13π/6, 17π/6) φ₁ = 2π + sin⁻¹(cosθ + 1/2) - θ φ₁ ∈ (2π/3, 5π/6), φ₂ ∈ (5π/6, 4π/3) ⇒ φ ∈ (2π/3, 4π/3) φ ∈ (2π/3, 4π/3) ⊂ (π/2, 4π/3) ⇒ (2) is satisfiable; (1), (3), (4) are not boxedCannot satisfy: (1), (3), (4)

Let $\theta, \varphi \in [0, 2\pi]$ be such that $2\cos\theta\,(1 - \sin\varphi) = \sin^{2}\theta \left(\tan\frac{\theta}{2} + \cot\frac{\theta}{2}\right)\cos\varphi - 1$ , $\tan(2\pi - \theta) > 0$ and $-1 < \sin\theta < -\frac{\sqrt{3}}{2}$ , then $\varphi$ cannot satisfy

Answers: A. $0 < \varphi < \frac{\pi}{2}$ · C. $\frac{4\pi}{3} < \varphi < \frac{3\pi}{2}$ · D. $\frac{3\pi}{2} < \varphi < 2\pi$

Trigonometric Functions Let

\theta, \varphi \in [0, 2\pi]

be such that

2\cos\theta\,(1 - \sin\varphi) = \sin^{2}\theta \left(\tan\frac{\theta}{2} + \cot\frac{\theta}{2}\right)\cos\varphi - 1

\tan(2\pi - \theta) > 0

and

-1 < \sin\theta < -\frac{\sqrt{3}}{2}

, then

\varphi

cannot satisfy

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Written solution

$2\cos\theta\,(1 - \sin\varphi) = \sin^2\theta\left(\tan\tfrac{\theta}{2} + \cot\tfrac{\theta}{2}\right)\cos\varphi - 1$

$\tan(2\pi - \theta) > 0$

$-1 < \sin\theta < -\tfrac{\sqrt{3}}{2}$

$\text{Find: interval } \varphi \text{ cannot satisfy}$

$\sin\theta = -\tfrac{\sqrt3}{2} \Rightarrow \theta = \tfrac{4\pi}{3},\ \tfrac{5\pi}{3}$

$\sin\theta = -1 \Rightarrow \theta = \tfrac{3\pi}{2}$

$\theta \in \left(\tfrac{4\pi}{3}, \tfrac{3\pi}{2}\right) \cup \left(\tfrac{3\pi}{2}, \tfrac{5\pi}{3}\right)$

$\tan(2\pi - \theta) = -\tan\theta > 0 \Rightarrow \tan\theta < 0$

$\theta \in \left(\tfrac{3\pi}{2}, \tfrac{5\pi}{3}\right)$

$\tan\tfrac{\theta}{2} + \cot\tfrac{\theta}{2} = \dfrac{2}{\sin\theta}$

$2\cos\theta(1-\sin\varphi) = 2\sin\theta\cos\varphi - 1$

$2\cos\theta + 1 = 2\sin\theta\cos\varphi + 2\cos\theta\sin\varphi$

$2\cos\theta + 1 = 2\sin(\theta + \varphi)$

$\sin(\theta + \varphi) = \cos\theta + \tfrac{1}{2}$

$\theta \in \left(\tfrac{3\pi}{2}, \tfrac{5\pi}{3}\right) \Rightarrow \cos\theta \in \left(0, \tfrac{1}{2}\right)$

$\sin(\theta + \varphi) \in \left(\tfrac{1}{2}, 1\right)$

$\theta + \varphi \in \left(\tfrac{3\pi}{2}, \tfrac{11\pi}{3}\right)$

$\sin s \in \left(\tfrac{1}{2}, 1\right) \Rightarrow s \in \left(\tfrac{13\pi}{6}, \tfrac{17\pi}{6}\right)$

$\theta + \varphi \in \left(\tfrac{13\pi}{6}, \tfrac{17\pi}{6}\right)$

$\varphi_1 = 2\pi + \sin^{-1}(\cos\theta + \tfrac{1}{2}) - \theta$

$\varphi_1 \in \left(\tfrac{2\pi}{3}, \tfrac{5\pi}{6}\right), \quad \varphi_2 \in \left(\tfrac{5\pi}{6}, \tfrac{4\pi}{3}\right)$

$\Rightarrow \varphi \in \left(\tfrac{2\pi}{3}, \tfrac{4\pi}{3}\right)$

$\varphi \in \left(\tfrac{2\pi}{3}, \tfrac{4\pi}{3}\right) \subset \left(\tfrac{\pi}{2}, \tfrac{4\pi}{3}\right)$

$\Rightarrow \text{(2) is satisfiable; (1), (3), (4) are not}$

$\boxed{\text{Cannot satisfy: (1), (3), (4)}}$

Written solution

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