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Mathematics Question on Trigonometry

Suppose θ∈[0,π4]\theta \in \left[ 0, \frac{\pi}{4} \right] is a solution of 4cos⁡θ−3sin⁡θ=14 \cos \theta - 3 \sin \theta = 1. Then cos⁡θ\cos \theta is equal to:

A

436−2\frac{4}{3\sqrt{6} - 2}

B

6−636−2\frac{6 - \sqrt{6}}{3\sqrt{6} - 2}

C

6+636+2\frac{6 + \sqrt{6}}{3\sqrt{6} + 2}

D

436+2\frac{4}{3\sqrt{6} + 2}

Answer

436−2\frac{4}{3\sqrt{6} - 2}

Explanation

Solution

Let 4(1−tan⁡2θ21+tan⁡2θ2)−3(2tan⁡θ21+tan⁡2θ2)=14 \left(\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\right) - 3 \left(\frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\right) = 1 Let tan⁡θ2=t\tan \frac{\theta}{2} = t, we have

4−4t2−6t1+t2=1\frac{4 - 4t^2 - 6t}{1 + t^2} = 1 Multiplying both sides by 1+t21 + t^2, we get: 4−4t2−6t=1+t24 - 4t^2 - 6t = 1 + t^2 Rearranging terms: 5t2+6t−3=05t^2 + 6t - 3 = 0

Solving this quadratic equation using the quadratic formula: t=−6±36−4×5×(−3)2×5t = \frac{-6 \pm \sqrt{36 - 4 \times 5 \times (-3)}}{2 \times 5} t=−6±9610t = \frac{-6 \pm \sqrt{96}}{10} t=−6±4610t = \frac{-6 \pm 4\sqrt{6}}{10} t=−3±265t = \frac{-3 \pm 2\sqrt{6}}{5}

Next, we find cos⁡θ\cos \theta: cos⁡θ=1−t21+t2\cos \theta = \frac{1 - t^2}{1 + t^2} Substituting t=−3+265t = \frac{-3 + 2\sqrt{6}}{5}, cos⁡θ=1−(−3+265)21+(−3+265)2\cos \theta = \frac{1 - \left(\frac{-3 + 2\sqrt{6}}{5}\right)^2}{1 + \left(\frac{-3 + 2\sqrt{6}}{5}\right)^2}

Calculating further, =1−(24+9−126)251+(24+9−126)25= \frac{1 - \frac{(24 + 9 - 12\sqrt{6})}{25}}{1 + \frac{(24 + 9 - 12\sqrt{6})}{25}} =25−(33−126)2525+(33−126)25= \frac{\frac{25 - (33 - 12\sqrt{6})}{25}}{\frac{25 + (33 - 12\sqrt{6})}{25}} =25−33+12625+33−126= \frac{25 - 33 + 12\sqrt{6}}{25 + 33 - 12\sqrt{6}} =−8+12658−126= \frac{-8 + 12\sqrt{6}}{58 - 12\sqrt{6}} =4−664−66= \frac{4 - 6\sqrt{6}}{4 - 6\sqrt{6}}
=−20025(4−66)= \frac{-200}{25(4 - 6\sqrt{6})} =−84−66= \frac{-8}{4 - 6\sqrt{6}} =436−2= \frac{4}{3\sqrt{6} - 2}