高専の数学解説

第12章 12.12

(1)

(1+1cos⁡θ)(1−cos⁡θ)=(cos⁡θ+1cos⁡θ)(1−cos⁡θ)=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=sin⁡θtan⁡θ\left(1+\frac{1}{\cos\theta}\right)(1-\cos\theta)=\left(\frac{\cos\theta+1}{\cos\theta}\right)(1-\cos\theta)=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}=\sin\theta\tan\theta

(2)

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

(3)

tan⁡θsin⁡θ−sin⁡θtan⁡θ=1cos⁡θ−cos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=sin⁡θtan⁡θ\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}=\sin\theta\tan\theta

解説: r31bn1z