题目
68. sin (15)^circ = ()-|||-A. dfrac (sqrt {2+sqrt {3)}}(2) B. sqrt (dfrac {2+sqrt {3)}(2)} C. sqrt (dfrac {2-sqrt {3)}(2)} D. dfrac (sqrt {2-sqrt {3)}}(2)
题目解答
答案
解析
步骤 1：使用和差公式
$\sin {15}^{\circ }=\sin ({45}^{\circ }-{30}^{\circ })$，根据和差公式，$\sin (A-B)=\sin A\cos B-\cos A\sin B$，代入$A=45^{\circ}$和$B=30^{\circ}$，得到$\sin {15}^{\circ }=\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ }$。
步骤 2：代入特殊角的三角函数值
$\sin {45}^{\circ }=\dfrac {\sqrt {2}}{2}$，$\cos {30}^{\circ }=\dfrac {\sqrt {3}}{2}$，$\cos {45}^{\circ }=\dfrac {\sqrt {2}}{2}$，$\sin {30}^{\circ }=\dfrac {1}{2}$，代入上式，得到$\sin {15}^{\circ }=\dfrac {\sqrt {2}}{2}\cdot \dfrac {\sqrt {3}}{2}-\dfrac {\sqrt {2}}{2}\cdot \dfrac {1}{2}$。
步骤 3：化简
$\sin {15}^{\circ }=\dfrac {\sqrt {6}}{4}-\dfrac {\sqrt {2}}{4}=\dfrac {\sqrt {6}-\sqrt {2}}{4}$。为了与选项匹配，我们进一步化简$\sqrt {6}-\sqrt {2}$，得到$\sqrt {6}-\sqrt {2}=\sqrt {8-4\sqrt {3}}=2\sqrt {2-\sqrt {3}}$，因此$\sin {15}^{\circ }=\dfrac {\sqrt {2-\sqrt {3}}}{2}$。
$\sin {15}^{\circ }=\sin ({45}^{\circ }-{30}^{\circ })$，根据和差公式，$\sin (A-B)=\sin A\cos B-\cos A\sin B$，代入$A=45^{\circ}$和$B=30^{\circ}$，得到$\sin {15}^{\circ }=\sin {45}^{\circ }\cos {30}^{\circ }-\cos {45}^{\circ }\sin {30}^{\circ }$。
步骤 2：代入特殊角的三角函数值
$\sin {45}^{\circ }=\dfrac {\sqrt {2}}{2}$，$\cos {30}^{\circ }=\dfrac {\sqrt {3}}{2}$，$\cos {45}^{\circ }=\dfrac {\sqrt {2}}{2}$，$\sin {30}^{\circ }=\dfrac {1}{2}$，代入上式，得到$\sin {15}^{\circ }=\dfrac {\sqrt {2}}{2}\cdot \dfrac {\sqrt {3}}{2}-\dfrac {\sqrt {2}}{2}\cdot \dfrac {1}{2}$。
步骤 3：化简
$\sin {15}^{\circ }=\dfrac {\sqrt {6}}{4}-\dfrac {\sqrt {2}}{4}=\dfrac {\sqrt {6}-\sqrt {2}}{4}$。为了与选项匹配，我们进一步化简$\sqrt {6}-\sqrt {2}$，得到$\sqrt {6}-\sqrt {2}=\sqrt {8-4\sqrt {3}}=2\sqrt {2-\sqrt {3}}$，因此$\sin {15}^{\circ }=\dfrac {\sqrt {2-\sqrt {3}}}{2}$。