高専の数学解説

第12章 12.13

sin⁡θ+cos⁡θ=a\sin\theta + \cos\theta = a

(sin⁡θ+cos⁡θ)2=a2(\sin\theta + \cos\theta)^2 = a^2

sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=a22sin⁡θcos⁡θ=a2−1sin⁡θcos⁡θ=12(a2−1)\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = a^2 \qquad 2\sin\theta\cos\theta = a^2 - 1 \qquad \sin\theta\cos\theta = \frac{1}{2}(a^2-1)

これから

sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)3−3sin⁡2θcos⁡θ−3sin⁡θcos⁡2θ\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)^3 - 3\sin^2\theta\cos\theta - 3\sin\theta\cos^2\theta

=(sin⁡θ+cos⁡θ)3−3sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)=a3−32(a2−1)a=a3−32a3+32a=−12a3+32a= (\sin\theta + \cos\theta)^3 - 3\sin\theta\cos\theta(\sin\theta + \cos\theta) = a^3 - \frac{3}{2}(a^2-1)a = a^3 - \frac{3}{2}a^3 + \frac{3}{2}a = -\frac{1}{2}a^3 + \frac{3}{2}a

解説: r31bn1z