高専の数学解説

第13章 13.15

(1)

(sin⁡θ+cos⁡θ)2=sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1+2sin⁡θcos⁡θ=1+sin⁡2θ(\sin\theta+\cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta = 1 + \sin2\theta

(2)

cos⁡4θ−sin⁡4θ=(cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)=cos⁡2θ−sin⁡2θ=12(1+cos⁡2θ)−12(1−cos⁡2θ)=cos⁡2θ\cos^4\theta - \sin^4\theta = (\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta) = \cos^2\theta - \sin^2\theta = \frac{1}{2}(1+\cos2\theta) - \frac{1}{2}(1-\cos2\theta) = \cos2\theta

(3)

1−cos⁡2θsin⁡2θ=1−cos⁡2θ+sin⁡2θ2sin⁡θcos⁡θ=sin⁡2θ+sin⁡2θ2sin⁡θcos⁡θ=sin⁡θcos⁡θ=tan⁡θ\frac{1-\cos2\theta}{\sin2\theta} = \frac{1-\cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin^2\theta+\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta

(4)

cos⁡2θ=cos⁡2θ−sin⁡2θ=cos⁡2θ−sin⁡2θ1=cos⁡2θ−sin⁡2θcos⁡2θ+sin⁡2θ=1−sin⁡2θcos⁡2θ1+sin⁡2θcos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos2\theta = \cos^2\theta - \sin^2\theta = \frac{\cos^2\theta-\sin^2\theta}{1} = \frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \frac{1-\dfrac{\sin^2\theta}{\cos^2\theta}}{1+\dfrac{\sin^2\theta}{\cos^2\theta}} = \frac{1-\tan^2\theta}{1+\tan^2\theta}

解説: r31bn1z