高専の数学解説

第12章 12.14

(1)

1+tan⁡θ1−tan⁡θ=2\frac{1+\tan\theta}{1-\tan\theta} = 2

1+tan⁡θ1−tan⁡θ=1+sin⁡θcos⁡θ1−sin⁡θcos⁡θ=cos⁡θ+sin⁡θcos⁡θ−sin⁡θ=2\frac{1+\tan\theta}{1-\tan\theta} = \frac{1+\dfrac{\sin\theta}{\cos\theta}}{1-\dfrac{\sin\theta}{\cos\theta}} = \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta} = 2

cos⁡θ=−3sin⁡θ=−31−cos⁡2θ\cos\theta = -3\sin\theta = -3\sqrt{1-\cos^2\theta}

cos⁡2θ=9(1−cos⁡2θ)\cos^2\theta = 9(1-\cos^2\theta)

10cos⁡2θ=9cos⁡2θ=910cos⁡θ=±31010\cos^2\theta = 9 \qquad \cos^2\theta = \frac{9}{10} \qquad \cos\theta = \pm\frac{3}{\sqrt{10}}

(2)

sin⁡θ+cos⁡θ=12(sin⁡θ+cos⁡θ)2=14sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=14sin⁡θcos⁡θ=−38\sin\theta+\cos\theta = \frac{1}{2} \qquad (\sin\theta+\cos\theta)^2 = \frac{1}{4} \qquad \sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta = \frac{1}{4} \qquad \sin\theta\cos\theta = -\frac{3}{8}

これから

tan⁡θ+1tan⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1−38=−83\tan\theta+\frac{1}{\tan\theta} = \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{-\frac{3}{8}} = -\frac{8}{3}

解説: r31bn1z