高専の数学解説

第13章 13.4

sin⁡x=13\sin x = \dfrac{1}{3} なので cos⁡x=83\cos x = \dfrac{\sqrt{8}}{3} tan⁡x=18\tan x = \dfrac{1}{\sqrt{8}} となる

(1)

sin⁡(x−π3)=sin⁡x⋅cos⁡π3−cos⁡x⋅sin⁡π3=13⋅12−83⋅32=16−266=1−266\sin\left(x-\frac{\pi}{3}\right) = \sin x \cdot \cos\frac{\pi}{3} - \cos x \cdot \sin\frac{\pi}{3} = \frac{1}{3}\cdot\frac{1}{2} - \frac{\sqrt{8}}{3}\cdot\frac{\sqrt{3}}{2} = \frac{1}{6} - \frac{2\sqrt{6}}{6} = \frac{1-2\sqrt{6}}{6}

(2)

cos⁡(x−π3)=cos⁡x⋅cos⁡π3+sin⁡x⋅sin⁡π3=83⋅12+13⋅32=22+36\cos\left(x-\frac{\pi}{3}\right) = \cos x \cdot \cos\frac{\pi}{3} + \sin x \cdot \sin\frac{\pi}{3} = \frac{\sqrt{8}}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{\sqrt{3}}{2} = \frac{2\sqrt{2}+\sqrt{3}}{6}

(3)

tan⁡(x−π3)=tan⁡x−tan⁡π31+tan⁡x⋅tan⁡π3=18−31+18⋅3=1−2622+3=(1−26)(22−3)8−3=82−935\tan\left(x-\frac{\pi}{3}\right) = \frac{\tan x - \tan\dfrac{\pi}{3}}{1+\tan x \cdot \tan\dfrac{\pi}{3}} = \frac{\dfrac{1}{\sqrt{8}}-\sqrt{3}}{1+\dfrac{1}{\sqrt{8}}\cdot\sqrt{3}} = \frac{1-2\sqrt{6}}{2\sqrt{2}+\sqrt{3}} = \frac{(1-2\sqrt{6})(2\sqrt{2}-\sqrt{3})}{8-3} = \frac{8\sqrt{2}-9\sqrt{3}}{5}

解説: r31bn1z