Question

довести тотожність................

Answer 1

$1)sin \alpha +cos \alpha = \sqrt2(cos \frac{ \pi }{4}- \alpha )}\\sin \alpha +cos \alpha = \sqrt2(cos \frac \pi 4cos \alpha +sin \frac \pi 4sin \alpha )\\sin \alpha +cos \alpha= \sqrt{2}( \frac{ \sqrt{2}}{2}cos \alpha + \frac{ \sqrt{2}}{2}sin \alpha )}\\sin \alpha +cos \alpha = \frac{( \sqrt2)^2}{2}sin\alpha +\frac{( \sqrt2)^2}{2}cos \alpha \\sin \alpha +cos \alpha = \frac22sin \alpha +\frac22cos \alpha \\sin \alpha +cos \alpha =sin \alpha +cos \alpha$
$2)sin \alpha -cos \alpha =- \sqrt2(cos \frac{ \pi }{4}+ \alpha )}\\sin \alpha -cos \alpha =- \sqrt2(cos \frac \pi 4cos \alpha -sin \frac \pi 4sin \alpha )\\sin \alpha -cos \alpha=- \sqrt{2}( \frac{ \sqrt{2}}{2}cos \alpha - \frac{ \sqrt{2}}{2}sin \alpha )}\\sin \alpha -cos \alpha = \frac{( \sqrt2)^2}{2}sin \alpha - \frac{( \sqrt2)^2}{2}cos \alpha \\sin \alpha -cos \alpha = \frac22sin \alpha - \frac22cos \alpha \\sin \alpha -cos \alpha =sin \alpha -cos \alpha$
$3) \frac{sin2 \alpha +sin6 \alpha }{cos2 \alpha +cos6 \alpha }=tg4 \alpha \\ \frac{2sin( \frac{2 \alpha +6 \alpha}{2} )cos(\frac{2 \alpha -6 \alpha}{2})}{2cos(\frac{2 \alpha +6 \alpha}{2})cos(\frac{2 \alpha -6 \alpha}{2})}=tg4 \alpha \\ \frac{2sin4 \alpha }{2cos4 \alpha }=tg4 \alpha \\ \frac{sin4 \alpha }{cos4 \alpha }=tg4 \alpha \\tg4 \alpha =tg4 \alpha$
$4) \frac{cos2 \alpha -cos4 \alpha }{cos2 \alpha +cos4 \alpha }=tg3 \alpha tg \alpha\\ \frac{-2sin( \frac{2 \alpha +4 \alpha }{2})sin( \frac{2 \alpha -4 \alpha }{2})}{2cos( \frac{2 \alpha +4 \alpha }{2})cos( \frac{2 \alpha -4 \alpha }{2})}=tg 3\alpha tg \alpha \\ \frac{2sin3 \alpha sin \alpha }{2cos3 \alpha cos \alpha }=tg3 \alpha tg \alpha \\ \frac{sin3 \alpha sin \alpha }{cos3 \alpha cos \alpha }=tg3 \alpha tg \alpha \\tg3 \alpha tg \alpha =tg3 \alpha tg \alpha$
$5)\frac{sin \alpha +sin5\alpha }{cos \alpha +cos5 \alpha }=tg3 \alpha \\ \frac{2sin( \frac{\alpha +5\alpha}{2} )cos(\frac{\alpha -5 \alpha}{2})}{2cos(\frac{ \alpha +5 \alpha}{2})cos(\frac{ \alpha -5\alpha}{2})}=tg3 \alpha \\ \frac{2sin3\alpha }{2cos3 \alpha }=tg3 \alpha \\ \frac{sin3\alpha }{cos3 \alpha }=tg3 \alpha \\tg3 \alpha =tg3\alpha$
$6) \frac{sin2 \alpha +sin \alpha }{sin2 \alpha -sin \alpha }= \frac{tg \frac{3 \alpha }{2}}{tg \frac{ \alpha }{2}}\\ \frac{2sin (\frac{2 \alpha + \alpha }{2})cos (\frac{2 \alpha - \alpha }{2})}{2sin(\frac{2 \alpha - \alpha }{2})cos(\frac{2 \alpha + \alpha }{2})}= \frac{tg \frac{3 \alpha }{2}}{tg \frac{ \alpha }{2}}\\ \frac{sin \frac{3 \alpha }{2}cos \frac{ \alpha }{2}}{cos \frac{3 \alpha }{2}sin \frac{ \alpha }{2}}=\frac{tg \frac{3 \alpha }{2}}{tg \frac{ \alpha }{2}}$
$\frac{tg \frac{3 \alpha }{2}}{tg \frac{ \alpha }{2}}=\frac{tg \frac{3 \alpha }{2}}{tg \frac{ \alpha }{2}}$
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