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MathHardAdv 2009 • Paper 2

Solutions to Trigonometric Summation Involving Cosecant Products

For 0<θ<π20<\theta<\frac{\pi}{2}, the solution(s) of
∑m=16cosec⁡(θ+(m−1)π4)cosec⁡(θ+mπ4)=42\sum_{m=1}^{6} \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}
is(are)

Options

A

π4\frac{\pi}{4}

B

π6\frac{\pi}{6}

C

π12\frac{\pi}{12}

Correct
D

5π12\frac{5 \pi}{12}

Correct

Topics & Concepts

Trigonometric FunctionsTrigonometric EquationsTrigonometric Ratios and Identities

Step-by-Step Solution

To find the solution(s) of the given equation for 0<θ<π20 < \theta < \frac{\pi}{2}:
∑m=16cosec⁡(θ+(m−1)π4)cosec⁡(θ+mπ4)=42\sum_{m=1}^{6} \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}

Let the general term be:
Tm=cosec⁡(θ+(m−1)π4)cosec⁡(θ+mπ4)=1sin⁡(θ+(m−1)π4)sin⁡(θ+mπ4)T_m = \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) = \frac{1}{\sin\left(\theta+\frac{(m-1) \pi}{4}\right) \sin\left(\theta+\frac{m \pi}{4}\right)}

Notice that the difference between the arguments is:
(θ+mπ4)−(θ+(m−1)π4)=π4\left(\theta + \frac{m\pi}{4}\right) - \left(\theta + \frac{(m-1)\pi}{4}\right) = \frac{\pi}{4}

Multiplying and dividing by sin⁡(π4)=12\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, we can express each term as:
sin⁡(π4)Tm=sin⁡[(θ+mπ4)−(θ+(m−1)π4)]sin⁡(θ+(m−1)π4)sin⁡(θ+mπ4)\sin\left(\frac{\pi}{4}\right) T_m = \frac{\sin\left[\left(\theta + \frac{m\pi}{4}\right) - \left(\theta + \frac{(m-1)\pi}{4}\right)\right]}{\sin\left(\theta+\frac{(m-1) \pi}{4}\right) \sin\left(\theta+\frac{m \pi}{4}\right)}

Using the identity sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A \cos B - \cos A \sin B:
12Tm=cot⁡(θ+(m−1)π4)−cot⁡(θ+mπ4)\frac{1}{\sqrt{2}} T_m = \cot\left(\theta+\frac{(m-1)\pi}{4}\right) - \cot\left(\theta+\frac{m\pi}{4}\right)
Tm=2[cot⁡(θ+(m−1)π4)−cot⁡(θ+mπ4)]T_m = \sqrt{2}\left[ \cot\left(\theta+\frac{(m-1)\pi}{4}\right) - \cot\left(\theta+\frac{m\pi}{4}\right) \right]

Summing from m=1m = 1 to 66, the terms telescope:
∑m=16Tm=2∑m=16[cot⁡(θ+(m−1)π4)−cot⁡(θ+mπ4)]\sum_{m=1}^{6} T_m = \sqrt{2} \sum_{m=1}^{6} \left[ \cot\left(\theta+\frac{(m-1)\pi}{4}\right) - \cot\left(\theta+\frac{m\pi}{4}\right) \right]
=2[cot⁡θ−cot⁡(θ+6π4)]= \sqrt{2} \left[ \cot\theta - \cot\left(\theta + \frac{6\pi}{4}\right) \right]
=2[cot⁡θ−cot⁡(3π2+θ)]= \sqrt{2} \left[ \cot\theta - \cot\left(\frac{3\pi}{2} + \theta\right) \right]

Since cot⁡(3π2+θ)=−tan⁡θ\cot\left(\frac{3\pi}{2} + \theta\right) = -\tan\theta, this simplifies to:
∑m=16Tm=2(cot⁡θ+tan⁡θ)\sum_{m=1}^{6} T_m = \sqrt{2} (\cot\theta + \tan\theta)

Setting this equal to 424\sqrt{2}:
2(cot⁡θ+tan⁡θ)=42\sqrt{2} (\cot\theta + \tan\theta) = 4\sqrt{2}
cot⁡θ+tan⁡θ=4\cot\theta + \tan\theta = 4

Rewriting in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta:
cos⁡θsin⁡θ+sin⁡θcos⁡θ=4  ⟹  cos⁡2θ+sin⁡2θsin⁡θcos⁡θ=4\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = 4 \implies \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = 4
1sin⁡θcos⁡θ=4  ⟹  2sin⁡2θ=4\frac{1}{\sin\theta\cos\theta} = 4 \implies \frac{2}{\sin 2\theta} = 4
sin⁡2θ=12\sin 2\theta = \frac{1}{2}

Since 0<θ<π20 < \theta < \frac{\pi}{2}, we have 0<2θ<π0 < 2\theta < \pi. Therefore:
2θ=π6or2θ=π−π6=5π62\theta = \frac{\pi}{6} \quad \text{or} \quad 2\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
θ=π12orθ=5π12\theta = \frac{\pi}{12} \quad \text{or} \quad \theta = \frac{5\pi}{12}

Thus, the correct options are C and D.

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