Question

14．已知：f（x）=$\left\{\begin{array}{l}{sinπx，x＜0}\\{f（x-1）+1，x≥0}\end{array}\right.$g（x）=$\left\{\begin{array}{l}{cosπx，x＜\frac{1}{2}}\\{g（x-1）-1，x≥\frac{1}{2}}\end{array}\right.$
求證：g（$\frac{1}{4}$）+f（$\frac{1}{3}$）+g（$\frac{5}{6}$）+f（$\frac{3}{4}$）=1．

Answer 1

分析 直接把x=$\frac{1}{4}$、$\frac{1}{3}$、$\frac{5}{6}$、$\frac{3}{4}$代入所對應(yīng)的分段函數(shù)，然后利用三角函數(shù)的誘導(dǎo)公式化簡求值．

解答 證明：∵f（x）=$\left\{\begin{array}{l}{sinπx，x＜0}\\{f（x-1）+1，x≥0}\end{array}\right.$，g（x）=$\left\{\begin{array}{l}{cosπx，x＜\frac{1}{2}}\\{g（x-1）-1，x≥\frac{1}{2}}\end{array}\right.$，
∴g（$\frac{1}{4}$）+f（$\frac{1}{3}$）+g（$\frac{5}{6}$）+f（$\frac{3}{4}$）=cos$\frac{π}{4}$+[f（-$\frac{2}{3}$）+1]+[g（-$\frac{1}{6}$）-1]+[f（-$\frac{1}{4}$）+1]
=$cos\frac{π}{4}$+[sin（-$\frac{2π}{3}$）+1]+[cos（-$\frac{π}{6}$）-1]+[sin（-$\frac{π}{4}$）+1]
=$\frac{\sqrt{2}}{2}$-$\frac{\sqrt{3}}{2}+1$+$\frac{\sqrt{3}}{2}-1$-$\frac{\sqrt{2}}{2}+1$=1．

點(diǎn)評 本題考查分段函數(shù)的應(yīng)用，考查了三角函數(shù)的誘導(dǎo)公式及三角函數(shù)值的求法，是中檔題．