高専の数学解説

第12章 12.3

(1)

sin⁡2θ1−cos⁡θ=1−cos⁡2θ1−cos⁡θ=(1−cos⁡θ)(1+cos⁡θ)1−cos⁡θ=1+cos⁡θ\frac{\sin^2\theta}{1-\cos\theta} = \frac{1-\cos^2\theta}{1-\cos\theta} = \frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta} = 1+\cos\theta

(2)

sin⁡4θ−cos⁡4θ=(sin⁡2θ+cos⁡2θ)(sin⁡2θ−cos⁡2θ)=sin⁡2θ−cos⁡2θ=sin⁡2θ−1+sin⁡2θ=2sin⁡2θ−1\sin^4\theta - \cos^4\theta = (\sin^2\theta+\cos^2\theta)(\sin^2\theta-\cos^2\theta) = \sin^2\theta-\cos^2\theta = \sin^2\theta-1+\sin^2\theta = 2\sin^2\theta-1

(3)

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

(4)

問題のtan⁡2θ\tan^2\thetaはtan⁡θ\tan\thetaとすれば両辺が等しくなる

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

=cos⁡θ1−sin⁡θ= \frac{\cos\theta}{1-\sin\theta}

(5)

tan⁡θ+cos⁡θ1+sin⁡θ=sin⁡θcos⁡θ+cos⁡θ(1−sin⁡θ)cos⁡2θ=sin⁡θcos⁡θ+cos⁡θ−cos⁡θsin⁡θcos⁡2θ=1cos⁡θ\tan\theta + \frac{\cos\theta}{1+\sin\theta} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta(1-\sin\theta)}{\cos^2\theta} = \frac{\sin\theta\cos\theta + \cos\theta - \cos\theta\sin\theta}{\cos^2\theta} = \frac{1}{\cos\theta}

解説: r31bn1z