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Algebra Difficulty 3.0 AMC 10/12 Find the answer

Given sin⁡(π+α)=−13\sin(\pi + \alpha) = -\frac{1}{3}, and α\alpha is an angle in the second quadrant, find the values of the following expressions:
(Ⅰ) cos⁡(2π−α)\cos(2\pi - \alpha);
(Ⅱ) tan⁡(α−7π)\tan(\alpha - 7\pi).

A number or a short expression. Spacing and $ signs are ignored.

Solutions — 2

Solution 1

(Ⅰ) Since sin⁡(π+α)=−sin⁡α=−13\sin(\pi + \alpha) = -\sin\alpha = -\frac{1}{3}, we have sin⁡α=13\sin\alpha = \frac{1}{3}.
As α\alpha is an angle in the second quadrant, we have cos⁡α=−223\cos\alpha = -\frac{2\sqrt{2}}{3}.
Therefore, cos⁡(2π−α)=cos⁡α=−223\cos(2\pi - \alpha) = \cos\alpha = -\frac{2\sqrt{2}}{3}.
So, the answer is −223\boxed{-\frac{2\sqrt{2}}{3}}.

(Ⅱ) tan⁡(α−7π)=tan⁡α=sin⁡αcos⁡α=−122=−24\tan(\alpha - 7\pi) = \tan\alpha = \frac{\sin\alpha}{\cos\alpha} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}.
Thus, the answer is −24\boxed{-\frac{\sqrt{2}}{4}}.

Solution 2

For (Ⅰ), since sin⁡(π+α)=−sin⁡α=−13\sin (\pi+\alpha)=-\sin \alpha=- \frac {1}{3}, we have sin⁡α=13\sin \alpha= \frac {1}{3}.
As α\alpha is an angle in the second quadrant, we get cos⁡α=−223\cos \alpha=- \frac {2 \sqrt {2}}{3}.
Therefore, cos⁡(2π−α)=cos⁡α=−223\cos (2\pi-\alpha)=\cos \alpha=- \frac {2 \sqrt {2}}{3}.

For (Ⅱ), tan⁡(α−7π)=tan⁡α=sin⁡αcos⁡α=−122=−24\tan (\alpha-7\pi)=\tan \alpha= \frac {\sin \alpha}{\cos \alpha }=- \frac {1}{2 \sqrt {2}}=- \frac { \sqrt {2}}{4}.

Thus, the answers are:
(Ⅰ) −223\boxed{- \frac {2 \sqrt {2}}{3}};
(Ⅱ) −24\boxed{- \frac { \sqrt {2}}{4}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.