第13章 13.26

sin3x+sin(x+π2)=2sin(2x+π4)cos(xπ4)=2sin(2x+π4)sin(x+π4)=3sin(x+π4)\sin 3x + \sin\left(x+\dfrac{\pi}{2}\right) = 2\sin\left(2x+\dfrac{\pi}{4}\right)\cos\left(x-\dfrac{\pi}{4}\right) = 2\sin\left(2x+\dfrac{\pi}{4}\right)\sin\left(x+\dfrac{\pi}{4}\right) = \sqrt{3}\sin\left(x+\dfrac{\pi}{4}\right)

sin(x+π4)=0\sin\left(x+\dfrac{\pi}{4}\right) = 0 これから π4+nπ-\dfrac{\pi}{4}+n\pi

sin(x+π4)0\sin\left(x+\dfrac{\pi}{4}\right) \neq 0 のとき

sin(2x+π4)=32\sin\left(2x+\dfrac{\pi}{4}\right) = \dfrac{\sqrt{3}}{2} これから π24+nπ\dfrac{\pi}{24}+n\pi524π+nπ\dfrac{5}{24}\pi+n\pi

解説: r31bn1z