第13章 13.8

(1)

sin⁡π9+sin⁡29π=2sin⁡π9+29π2cos⁡(π9−29π)2=2sin⁡π6cos⁡π18\sin\frac{\pi}{9}+\sin\frac{2}{9}\pi=2\sin\frac{\dfrac{\pi}{9}+\dfrac{2}{9}\pi}{2}\cos\frac{\left(\dfrac{\pi}{9}-\dfrac{2}{9}\pi\right)}{2}=2\sin\frac{\pi}{6}\cos\frac{\pi}{18}

(2)

cos⁡29π+cos⁡79π=2cos⁡29π+79π2cos⁡29π−79π2=2cos⁡π2cos⁡518π\cos\frac{2}{9}\pi+\cos\frac{7}{9}\pi=2\cos\frac{\dfrac{2}{9}\pi+\dfrac{7}{9}\pi}{2}\cos\frac{\dfrac{2}{9}\pi-\dfrac{7}{9}\pi}{2}=2\cos\frac{\pi}{2}\cos\frac{5}{18}\pi

(3)

cos⁡34π−cos⁡23π=−2sin⁡34π+23π2sin⁡34π−23π2=−2sin⁡1724πsin⁡π24\cos\frac{3}{4}\pi-\cos\frac{2}{3}\pi=-2\sin\frac{\dfrac{3}{4}\pi+\dfrac{2}{3}\pi}{2}\sin\frac{\dfrac{3}{4}\pi-\dfrac{2}{3}\pi}{2}=-2\sin\frac{17}{24}\pi\sin\frac{\pi}{24}

(4)

sin⁡8x−sin⁡4x=2cos⁡8x+4x2sin⁡8x−4x2=2cos⁡6x⋅sin⁡2x\sin8x-\sin4x=2\cos\frac{8x+4x}{2}\sin\frac{8x-4x}{2}=2\cos6x\cdot\sin2x

(5)

cos⁡2x+cos⁡4x=2cos⁡2x+4x2cos⁡2x−4x2=2cos⁡3x⋅cos⁡x\cos2x+\cos4x=2\cos\frac{2x+4x}{2}\cos\frac{2x-4x}{2}=2\cos3x\cdot\cos x

(6)

cos⁡6x−cos⁡2x=−2sin⁡6x+2x2⋅sin⁡6x−2x2=−2sin⁡4x⋅sin⁡2x\cos6x-\cos2x=-2\sin\frac{6x+2x}{2}\cdot\sin\frac{6x-2x}{2}=-2\sin4x\cdot\sin2x

解説: r31bn1z