计算下列式子的值
\( \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) \)
已知
\( \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) \)
要求
我们需要计算\( \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) \)的值。
解答:
我们知道:
$\sin\ (90^{\circ}- \theta) = \cos\ \theta$
$\tan\ (90^{\circ}- \theta) = \cot\ \theta$
因此:
$\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)=\left(\frac{\cot 40^{\circ} }{\tan( 90^{\circ}-40^{\circ} )}\right) -\frac{1}{2}\left(\frac{\cos 35^{\circ} }{\sin( 90^{\circ} -35^{\circ} )}\right)$
$=\left(\frac{\cot 40^{\circ} }{\cot 40^{\circ} }\right) -\frac{1}{2}\left(\frac{\cos 35^{\circ} }{\cos 35^{\circ} }\right)$
$=1-\frac{1}{2}( 1)$
$=\frac{2( 1)-1}{2}$
$=\frac{1}{2}$
因此,$\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)=\frac{1}{2}$.