计算下列式子的值
\( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} \)
已知
\( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} \)
要求
我们需要计算\( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} \)的值。
解:
我们知道:
$\sin\ (90^{\circ}- \theta) = \cos\ \theta$
因此:
$\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}=\left(\frac{\sin( 90^{\circ} -41^{\circ} )}{\cos 41^{\circ} }\right)^{2} +\left(\frac{\cos 41^{\circ} }{\sin( 90^{\circ} -41^{\circ} )}\right)^{2}$
$=\left(\frac{\cos 41^{\circ}}{\cos 41^{\circ} }\right)^{2} +\left(\frac{\cos 41^{\circ} }{\cos 41^{\circ} }\right)^{2}$
$=( 1)^{2} +( 1)^{2}$
$=1+1$
$=2$
因此,$\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}=2$。