Question

докажите тождество sinx-cosx=-√2cos(π/4+х)

Answer 1

$sinx-cosx = -\sqrt{2}(cos \frac{ \pi }{4}+x) \\ -\sqrt{2}cos( \frac{ \pi }{4}+x) =$
$\sqrt{2}(cos \frac{ \pi }{4}*cosx-sin \frac{ \pi }{4}*sinx) = - \sqrt{2} ( \frac{ \sqrt{2} }{2}*cosx- \frac{ \sqrt{2} }{2}sinx) =$ $-\sqrt{2}* \frac{ \sqrt{2} }{2}(cosx-sinx) = -\frac{2}{2}(cosx - sinx)=-(cosx - sin x)=$ $sinx-cosx$