求解三角方程 sinx／sin(80°-x)=(sin30°sin50°)／(sin80°sin80°)

popo987654

请问如何用三角恒等变换求出 x

luyuanhong

x=20°

王守恩

求解三角方程 sinx／sin(80°-x)=(sin30°sin50°)／(sin80°sin80°)

\(1=\frac{\sin x\sin80^\circ\sin80^\circ}{\sin(80^\circ-x)\sin30^\circ\sin50^\circ}=\frac{\sin x(2\sin40^\circ\cos40^\circ)\sin80^\circ}{\sin(80^\circ-x)\sin30^\circ\cos40^\circ}=\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\sin60^\circ\sin30^\circ\sin30^\circ}\)

\(瞪眼!\ \ \ x=20^\circ\)

\(\sin20^\circ\sin40^\circ\sin80^\circ\)

\(=(\sin40^\circ\sin20^\circ)\sin80^\circ\)

\(=\sin30^\circ(\cos(40-20)^\circ-\cos(40+20)^\circ)\sin80^\circ\)

\(=\sin30^\circ\cos20^\circ\sin80^\circ-\sin30^\circ\cos60^\circ\sin80^\circ\)

\(=\sin30^\circ(\sin80^\circ\cos20^\circ)-\sin30^\circ\cos60^\circ\sin80^\circ\)

\(=\sin30^\circ\sin30^\circ(\sin(80+20)^\circ+\sin(80-20)^\circ)-\sin30^\circ\cos60^\circ\sin80^\circ\)

\(=\sin30^\circ\sin30^\circ\sin100^\circ+\sin30^\circ\sin30^\circ\sin60^\circ-\sin30^\circ\cos60^\circ\sin80^\circ\)

\(=\sin30^\circ\sin30^\circ\sin60^\circ\)