Question

4．對(duì)任意x，y∈R，恒有$sinx+cosy=2sin（\frac{x-y}{2}+\frac{π}{4}）cos（\frac{x+y}{2}-\frac{π}{4}）$，則$sin\frac{7π}{24}cos\frac{13π}{24}$等于（　�。�

• A． $\frac{{1+\sqrt{2}}}{4}$
• B． $\frac{{1-\sqrt{2}}}{4}$
• C． $\frac{{\sqrt{3}+\sqrt{2}}}{4}$
• D． $\frac{{\sqrt{3}-\sqrt{2}}}{4}$

Answer 1

分析 根據(jù)式子$\left\{\begin{array}{l}{\frac{x-y}{2}+\frac{π}{4}=\frac{7π}{24}}\\{\frac{x+y}{2}-\frac{π}{4}=\frac{13π}{24}}\end{array}\right.$，解方程組得x、y的值，再代入已知等式即可求值．

解答 解：由方程組$\left\{\begin{array}{l}{\frac{x-y}{2}+\frac{π}{4}=\frac{7π}{24}}\\{\frac{x+y}{2}-\frac{π}{4}=\frac{13π}{24}}\end{array}\right.$，解得，$\left\{\begin{array}{l}{x=\frac{5π}{6}}\\{y=\frac{3π}{4}}\end{array}\right.$．
∴$sin\frac{7π}{24}cos\frac{13π}{24}$=$\frac{1}{2}$（sin$\frac{5π}{6}$+cos$\frac{3π}{4}$）=$\frac{1}{2}（\frac{1}{2}-\frac{\sqrt{2}}{2}）=\frac{1-\sqrt{2}}{4}$．
故選：B．

點(diǎn)評(píng) 本題考查三角函數(shù)公式的應(yīng)用：求值．根據(jù)式子先求出x，y是關(guān)鍵，屬于中檔題．