JEE Challenger
More from Trigonometric Functions
MathHardAdv 2012 • Paper 1

Range of Angle Phi in Combined Trigonometric Relation

Let θ,φ∈[0,2π]\theta, \varphi \in [0, 2\pi] be such that 2cos⁡θ(1−sin⁡φ)=sin⁡2θ(tan⁡θ2+cot⁡θ2)cos⁡φ−1,2\cos\theta(1 - \sin\varphi) = \sin^2\theta\left(\tan\frac{\theta}{2} + \cot\frac{\theta}{2}\right)\cos\varphi - 1, with tan⁡(2π−θ)>0\tan(2\pi - \theta) > 0 and −1<sin⁡θ<−32-1 < \sin\theta < -\frac{\sqrt{3}}{2}.

Then φ\varphi cannot satisfy which of the following conditions?

Options

A

0<φ<π20 < \varphi < \frac{\pi}{2}

Correct
B

π2<φ<4π3\frac{\pi}{2} < \varphi < \frac{4\pi}{3}

C

4π3<φ<3π2\frac{4\pi}{3} < \varphi < \frac{3\pi}{2}

Correct
D

3π2<φ<2π\frac{3\pi}{2} < \varphi < 2\pi

Correct

Topics & Concepts

Trigonometric FunctionsTrigonometric EquationsTrigonometric Ratios and Identities

Step-by-Step Solution

To find the range of φ\varphi, we start by simplifying the given trigonometric equation: 2cos⁡θ(1−sin⁡φ)=sin⁡2θ(tan⁡θ2+cot⁡θ2)cos⁡φ−12\cos\theta(1 - \sin\varphi) = \sin^2\theta\left(\tan\frac{\theta}{2} + \cot\frac{\theta}{2}\right)\cos\varphi - 1

Using the identity: tan⁡θ2+cot⁡θ2=sin⁡θ2cos⁡θ2+cos⁡θ2sin⁡θ2=sin⁡2θ2+cos⁡2θ2sin⁡θ2cos⁡θ2=112sin⁡θ=2sin⁡θ\tan\frac{\theta}{2} + \cot\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} + \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \frac{\sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{1}{\frac{1}{2}\sin\theta} = \frac{2}{\sin\theta}

Substituting this back into the equation: 2cos⁡θ(1−sin⁡φ)=sin⁡2θ(2sin⁡θ)cos⁡φ−12\cos\theta(1 - \sin\varphi) = \sin^2\theta\left(\frac{2}{\sin\theta}\right)\cos\varphi - 1 2cos⁡θ−2cos⁡θsin⁡φ=2sin⁡θcos⁡φ−12\cos\theta - 2\cos\theta\sin\varphi = 2\sin\theta\cos\varphi - 1

Rearranging the terms: 2sin⁡θcos⁡φ+2cos⁡θsin⁡φ=2cos⁡θ+12\sin\theta\cos\varphi + 2\cos\theta\sin\varphi = 2\cos\theta + 1 2sin⁡(θ+φ)=2cos⁡θ+12\sin(\theta + \varphi) = 2\cos\theta + 1 sin⁡(θ+φ)=cos⁡θ+12\sin(\theta + \varphi) = \cos\theta + \frac{1}{2}

Next, we analyze the given constraints on θ∈[0,2π]\theta \in [0, 2\pi]:

  1. tan⁡(2π−θ)>0  ⟹  −tan⁡θ>0  ⟹  tan⁡θ<0\tan(2\pi - \theta) > 0 \implies -\tan\theta > 0 \implies \tan\theta < 0.
  2. −1<sin⁡θ<−32-1 < \sin\theta < -\frac{\sqrt{3}}{2}, which implies sin⁡θ<0\sin\theta < 0.

Since sin⁡θ<0\sin\theta < 0 and tan⁡θ<0\tan\theta < 0, θ\theta must lie in the fourth quadrant, i.e., θ∈(3π2,2π)\theta \in \left(\frac{3\pi}{2}, 2\pi\right).

Within the fourth quadrant, the condition −1<sin⁡θ<−32-1 < \sin\theta < -\frac{\sqrt{3}}{2} gives: θ∈(3π2,5π3)\theta \in \left(\frac{3\pi}{2}, \frac{5\pi}{3}\right)

For θ∈(3π2,5π3)\theta \in \left(\frac{3\pi}{2}, \frac{5\pi}{3}\right), the cosine function is strictly increasing, so: cos⁡(3π2)<cos⁡θ<cos⁡(5π3)  ⟹  0<cos⁡θ<12\cos\left(\frac{3\pi}{2}\right) < \cos\theta < \cos\left(\frac{5\pi}{3}\right) \implies 0 < \cos\theta < \frac{1}{2}

Substituting this into the expression for sin⁡(θ+φ)\sin(\theta + \varphi): 12<sin⁡(θ+φ)<1\frac{1}{2} < \sin(\theta + \varphi) < 1

Since θ∈(3π2,5π3)\theta \in \left(\frac{3\pi}{2}, \frac{5\pi}{3}\right) and φ∈[0,2π]\varphi \in [0, 2\pi], we have: θ+φ∈(3π2,11π3)\theta + \varphi \in \left(\frac{3\pi}{2}, \frac{11\pi}{3}\right)

Within this interval, the inequality 12<sin⁡(θ+φ)<1\frac{1}{2} < \sin(\theta + \varphi) < 1 requires: θ+φ∈(2π+π6,2π+5π6)∖{2π+π2}=(13π6,17π6)∖{5π2}\theta + \varphi \in \left(2\pi + \frac{\pi}{6}, 2\pi + \frac{5\pi}{6}\right) \setminus \left\{2\pi + \frac{\pi}{2}\right\} = \left(\frac{13\pi}{6}, \frac{17\pi}{6}\right) \setminus \left\{\frac{5\pi}{2}\right\}

Thus, we have: 13π6<θ+φ<17π6\frac{13\pi}{6} < \theta + \varphi < \frac{17\pi}{6}

Using the bounds on θ\theta: 3π2<θ<5π3\frac{3\pi}{2} < \theta < \frac{5\pi}{3} −5π3<−θ<−3π2-\frac{5\pi}{3} < -\theta < -\frac{3\pi}{2}

Adding these inequalities to bound φ=(θ+φ)−θ\varphi = (\theta + \varphi) - \theta: φ>13π6−5π3=13π−10π6=3π6=π2\varphi > \frac{13\pi}{6} - \frac{5\pi}{3} = \frac{13\pi - 10\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} φ<17π6−3π2=17π−9π6=8π6=4π3\varphi < \frac{17\pi}{6} - \frac{3\pi}{2} = \frac{17\pi - 9\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}

Hence, φ\varphi must satisfy: φ∈(π2,4π3)\varphi \in \left(\frac{\pi}{2}, \frac{4\pi}{3}\right)

Therefore, φ\varphi cannot satisfy the conditions given in options (A), (C), and (D).

The correct options are A, C, D.

PreviousNext