第12章 12.6

(1)

(sin⁡θ+cos⁡θ)2=sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1+2sin⁡θcos⁡θ=(2)2(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta = (\sqrt{2})^2

sin⁡θcos⁡θ=12\sin\theta\cos\theta = \frac{1}{2}

(2)

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

解説: r31bn1z