Maths Olympiad Prep

Track / Stage 3 / 65 of 260 #65 of 1964

Problem 65

AMC 10/12, early questions
Algebra Difficulty 3.1 Find the answer

Given tan⁡2θ=−22\tan 2\theta = -2\sqrt{2}, and π<2θ<2π\pi < 2\theta < 2\pi.

(Ⅰ) Find the value of tan⁡θ\tan \theta;
(Ⅱ) Calculate the value of 2cos⁡2θ2−sin⁡θ−12sin⁡(θ+π4)\frac{2\cos^2 \frac{\theta}{2} - \sin \theta - 1}{\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

(1) Since tan⁡2θ=2tan⁡θ1−tan⁡2θ=−22\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = -2\sqrt{2},
we have tan⁡θ=−22\tan \theta = -\frac{\sqrt{2}}{2} or tan⁡θ=2\tan \theta = \sqrt{2}.
Given π<2θ<2π\pi < 2\theta < 2\pi, it follows that π2<θ<π\frac{\pi}{2} < \theta < \pi,
thus tan⁡θ=−22\tan \theta = -\frac{\sqrt{2}}{2}.

(2) The original expression can be simplified to 1+cos⁡θ−sin⁡θ−1sin⁡θ+cos⁡θ=1−tan⁡θ1+tan⁡θ=1−(−22)1+(−22)=3+22\frac{1 + \cos \theta - \sin \theta - 1}{\sin \theta + \cos \theta} = \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{1 - (-\frac{\sqrt{2}}{2})}{1 + (-\frac{\sqrt{2}}{2})} = 3 + 2\sqrt{2}.

Therefore, the answers are:
(Ⅰ) −22\boxed{-\frac{\sqrt{2}}{2}}
(Ⅱ) 3+22\boxed{3 + 2\sqrt{2}}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.