1.次の値を求めなさい。
(1)\(\sin15°\cos105°\)
\(\displaystyle =\frac{1}{2}\{\sin(15°+105°)+\sin(15°-105°)\}\)
\(\displaystyle =\frac{1}{2}\{\sin120°+\sin(-90°)\}\)
\(\displaystyle =\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1\right)\)
\(\displaystyle =\frac{\sqrt{3}-2}{4}\)
(2)\(\sin15°\sin75°\)
\(\displaystyle =-\frac{1}{2}\{\cos(15°+75°)-\cos(15°-75°)\}\)
\(\displaystyle =-\frac{1}{2}\{\cos90°-\cos(-60°)\}\)
\(\displaystyle =-\frac{1}{2}\left(0-\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\)
(3)\(\sin105°\cos15°\)
\(\displaystyle =\frac{1}{2}\{\sin(105°+15°)+\sin(105°-15°)\}\)
\(\displaystyle =\frac{1}{2}(\sin120°+\sin90°)\)
\(\displaystyle =\frac{1}{2}\left(\frac{\sqrt{3}}{2}+1\right)\)
\(\displaystyle =\frac{\sqrt{3}+2}{4}\)
(4)\(\cos15°\cos75°\)
\(\displaystyle =\frac{1}{2}\{\cos(15°+75°)+\cos(15°-75°)\}\)
\(\displaystyle =\frac{1}{2}\{(\cos90°+\cos(-60°)\}\)
\(\displaystyle =\frac{1}{2}\left(0+\frac{1}{2}\right)\)
\(\displaystyle =\frac{1}{4}\)
(5)\(\sin15°-\sin105°\)
\(\displaystyle =2\cos\frac{15°+105°}{2}\sin\frac{15°-105°}{2}\)
\(=2\cos60°\sin(-45°)\)
\(\displaystyle =2・\frac{1}{2}・\left(-\frac{1}{\sqrt{2}}\right)\)
\(\displaystyle =-\frac{\sqrt{2}}{2}\)
(6)\(\cos15°+\cos75°\)
\(\displaystyle =2\cos\frac{15°+75°}{2}\cos\frac{15°-75°}{2}\)
\(=2\cos45°\cos(-30°)\)
\(\displaystyle =2・\frac{1}{\sqrt{2}}・\frac{\sqrt{3}}{2}\)
\(\displaystyle =\frac{\sqrt{6}}{2}\)
(7)\(\sin15°+\sin75°\)
\(\displaystyle =2\sin\frac{15°+75°}{2}\cos\frac{15°-75°}{2}\)
\(=2\sin45°\cos(-30°)\)
\(\displaystyle =2・\frac{1}{\sqrt{2}}・\frac{\sqrt{3}}{2}\)
\(\displaystyle =\frac{\sqrt{6}}{2}\)
(8)\(\cos15°-\sin105°\)
\(\displaystyle =-2\sin\frac{15°+105°}{2}\sin\frac{15°-105°}{2}\)
\(=-2\sin60°\sin(-45°)\)
\(\displaystyle =-2・\frac{\sqrt{3}}{2}・\left(-\frac{1}{\sqrt{2}}\right)\)
\(\displaystyle =\frac{\sqrt{6}}{2}\)
2.次の関数の最大値と最小値を求めなさい。ただし、\(0\leqq x\leqq2\pi\)とする。
(1)\(y=\sin x\cos x-\cos^2x\)
\(\displaystyle y=\frac{\sin2x}{2}-\frac{1+\cos2x}{2}\)
\(\displaystyle \ \ =\frac{1}{2}(\sin2x-\cos2x)-\frac{1}{2}\)
\(\displaystyle \ \ =\frac{\sqrt{2}}{2}\sin\left(2x-\frac{\pi}{4}\right)-\frac{1}{2}\)
\(\displaystyle -1\leqq\sin\left(2x-\frac{\pi}{4}\right)\leqq1\)より、
\(\displaystyle \sin\left(2x-\frac{\pi}{4}\right)=1\)のとき、最大値\(\displaystyle \frac{\sqrt{2}-1}{2}\)
\(\displaystyle x=\frac{3\pi}{8},\frac{11\pi}{8}\)
\(\displaystyle \sin\left(2x-\frac{\pi}{4}\right)=-1\)のとき、最小値\(\displaystyle -\frac{\sqrt{2}+1}{2}\)
\(\displaystyle x=\frac{7\pi}{8},\frac{15\pi}{8}\)
よって、
最大値は\(\displaystyle \frac{\sqrt{2}-1}{2}\)(\(\displaystyle x=\frac{3\pi}{8},\frac{11\pi}{8}\)のとき)
最小値は\(\displaystyle -\frac{\sqrt{2}+1}{2}\)(\(\displaystyle x=\frac{7\pi}{8},\frac{15\pi}{8}\)のとき)
(2)\(y=\sin x\cos x-\sqrt{3}\sin^2x\)
\(\displaystyle y=\frac{\sin2x}{2}-\sqrt{3}\frac{1-\cos2x}{2}\)
\(\displaystyle \ \ =\frac{1}{2}(\sin2x+\sqrt{3}\cos2x)-\frac{\sqrt{3}}{2}\)
\(\displaystyle \ \ =\sin\left(2x+\frac{\pi}{3}\right)-\frac{\sqrt{3}}{2}\)
\(\displaystyle -1\leqq\sin\left(2x+\frac{\pi}{3}\right)\leqq1\)より、
\(\displaystyle \sin\left(2x+\frac{\pi}{3}\right)=1\)のとき、最大値\(\displaystyle \frac{2-\sqrt{3}}{2}\)
\(\displaystyle x=\frac{\pi}{12},\frac{13\pi}{12}\)
\(\displaystyle \sin\left(2x+\frac{\pi}{3}\right)=-1\)のとき、最小値\(\displaystyle -\frac{2+\sqrt{3}}{2}\)
\(\displaystyle x=\frac{7\pi}{12},\frac{19\pi}{12}\)
よって、
最大値は\(\displaystyle \frac{2-\sqrt{3}}{2}\)(\(\displaystyle x=\frac{\pi}{12},\frac{13\pi}{12}\)のとき)
最小値は\(\displaystyle -\frac{2+\sqrt{3}}{2}\)(\(\displaystyle x=\frac{7\pi}{12},\frac{19\pi}{12}\)のとき)