Question

Чему равно
$sin(\pi/6 + \alpha) + sin(\pi/6 - \alpha) - cos(\alpha)$

Answer 1

$sin( \frac{ \pi }{6} + \alpha )+ sin( \frac{ \pi }{6} - a) - cos \alpha = 2*sin( \frac{ \frac{ \pi }{6}+ \alpha + \frac{ \pi }{6} - a }{2} ) * cos( \frac{ \frac{ \pi }{6} + \alpha - \frac{ \pi }{6} + \alpha }{2}) -$$- cos \ alpha = 2*sin( \frac{ \frac{ \pi }{3} }{2} )*cos( \frac{2 \alpha }{2} ) - cos \alpha= 2*sin \frac{ \pi }{6} *cos \alpha - cos \alpha = 2* \frac{1}{2} *cos \alpha - cos \alpha= 0$

Answer 2

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