第12章 12.6

(1)

(sinθ+cosθ)2=sin2θ+2sinθcosθ+cos2θ=1+2sinθcosθ=(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

sinθcosθ=12\sin\theta\cos\theta = \frac{1}{2}

(2)

sin4θ+cos4θ=(sin2θ+cos2θ)22sin2θcos2θ=12(sinθcosθ)2=12(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