高専の数学解説

第13章 13.21

(1)

sin⁡x+sin⁡y=2sin⁡x+y2cos⁡x−y2=2sin⁡π6cos⁡(π6−y)=cos⁡(π6−y)\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} = 2\sin\frac{\pi}{6}\cos\left(\frac{\pi}{6}-y\right) = \cos\left(\frac{\pi}{6}-y\right)

これはcos⁡(−y)\cos(-y)が−π6-\dfrac{\pi}{6}平行移動しただけのcos⁡\cos関数なので

最大値はcos⁡(π6−y)=1\cos\left(\dfrac{\pi}{6}-y\right)=1のときで11

最小値はcos⁡(π6−y)=−1\cos\left(\dfrac{\pi}{6}-y\right)=-1のときで−1-1

(2)

sin⁡x+cos⁡y=sin⁡(π3−y)+cos⁡y=sin⁡π3cos⁡y−cos⁡π3sin⁡y+cos⁡y=32cos⁡y−12sin⁡y+cos⁡y\sin x + \cos y = \sin\left(\frac{\pi}{3}-y\right) + \cos y = \sin\frac{\pi}{3}\cos y - \cos\frac{\pi}{3}\sin y + \cos y = \frac{\sqrt3}{2}\cos y - \frac12\sin y + \cos y

=−12sin⁡y+3+22cos⁡y=(−12)2+(3+22)2sin⁡(y+α)=6+43+22sin⁡(y+α)= -\frac12\sin y + \frac{\sqrt3+2}{2}\cos y = \sqrt{\left(-\frac12\right)^2+\left(\frac{\sqrt3+2}{2}\right)^2}\sin(y+\alpha) = \frac{\sqrt{6+4\sqrt3+2}}{2}\sin(y+\alpha)

=(6+2)22sin⁡(y+α)=6+22sin⁡(y+α)= \frac{\sqrt{\left(\sqrt6+\sqrt2\right)^2}}{2}\sin(y+\alpha) = \frac{\sqrt6+\sqrt2}{2}\sin(y+\alpha)

最大値はsin⁡(y+α)=1\sin(y+\alpha)=1のときで 6+22\dfrac{\sqrt6+\sqrt2}{2}

最小値はsin⁡(y+α)=−1\sin(y+\alpha)=-1のときで −6+22-\dfrac{\sqrt6+\sqrt2}{2}

(3)

sin⁡xcos⁡y=12{sin⁡(x+y)+sin⁡(x−y)}=12{sin⁡π3+sin⁡(π3−2y)}=34+12sin⁡(π3−2y)\sin x\cos y = \frac12\{\sin(x+y)+\sin(x-y)\} = \frac12\left\{\sin\frac{\pi}{3}+\sin\left(\frac{\pi}{3}-2y\right)\right\} = \frac{\sqrt3}{4}+\frac12\sin\left(\frac{\pi}{3}-2y\right)

最大値はsin⁡(π3−2y)=1\sin\left(\dfrac{\pi}{3}-2y\right)=1のときで 3+24\dfrac{\sqrt3+2}{4}

最小値はsin⁡(π3−2y)=−1\sin\left(\dfrac{\pi}{3}-2y\right)=-1のときで 3−24\dfrac{\sqrt3-2}{4}

(4)

cos⁡xcos⁡y=12{cos⁡(x+y)+cos⁡(x−y)}=12{cos⁡π3+cos⁡(π3−2y)}=14+12cos⁡(π3−2y)\cos x \cos y = \frac{1}{2}\{\cos(x+y) + \cos(x-y)\} = \frac{1}{2}\left\{\cos\frac{\pi}{3} + \cos\left(\frac{\pi}{3}-2y\right)\right\} = \frac{1}{4} + \frac{1}{2}\cos\left(\frac{\pi}{3}-2y\right)

最大値はcos⁡(π3−2y)=1\cos\left(\dfrac{\pi}{3}-2y\right) = 1のときで 34\dfrac{3}{4}

最小値はcos⁡(π3−2y)=−1\cos\left(\dfrac{\pi}{3}-2y\right) = -1のときで −14-\dfrac{1}{4}

(5)

sin⁡2x+sin⁡2y=12{1−cos⁡2x+1−cos⁡2y}=1−12{cos⁡2x+cos⁡2y}=1−cos⁡(x+y)cos⁡(x−y)\sin^2 x + \sin^2 y = \frac{1}{2}\{1-\cos 2x + 1 - \cos 2y\} = 1 - \frac{1}{2}\{\cos 2x + \cos 2y\} = 1 - \cos(x+y)\cos(x-y)

=1−cos⁡π3cos⁡(π3−2y)=1−12cos⁡(π3−2y)= 1 - \cos\frac{\pi}{3}\cos\left(\frac{\pi}{3}-2y\right) = 1 - \frac{1}{2}\cos\left(\frac{\pi}{3}-2y\right)

最大値はcos⁡(π3−2y)=1\cos\left(\dfrac{\pi}{3}-2y\right) = 1のときで 32\dfrac{3}{2}

最小値はcos⁡(π3−2y)=−1\cos\left(\dfrac{\pi}{3}-2y\right) = -1のときで12\dfrac{1}{2}

(6)

sin⁡2x+cos⁡2y=12{{1−cos⁡2x+1+cos⁡2y}=1−12{cos⁡2x−cos⁡2y}=1+sin⁡(x+y)sin⁡(x−y)\sin^2 x + \cos^2 y = \frac{1}{2}\{\{1-\cos 2x + 1 + \cos 2y\} = 1 - \frac{1}{2}\{\cos 2x - \cos 2y\} = 1 + \sin(x+y)\sin(x-y)

=1+32sin⁡(π3−2y)= 1 + \frac{\sqrt{3}}{2}\sin\left(\frac{\pi}{3}-2y\right)

最大値はsin⁡(π3−2y)=1\sin\left(\dfrac{\pi}{3}-2y\right) = 1のときで2+32\dfrac{2+\sqrt{3}}{2}

最小値はsin⁡(π3−2y)=−1\sin\left(\dfrac{\pi}{3}-2y\right) = -1のときで2−32\dfrac{2-\sqrt{3}}{2}

解説: r31bn1z