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July 2022 1 7 Report

Докажите тождества

а) [tex] \frac{cos2a}{1+sin2a} = \frac{1-tga}{1+ tga} [/tex]

б)[tex]2cos(45 ^{\circ} +a)*cos(45 ^{\circ} - a) = cos 2a[/tex]

Answers & Comments

Apofeoz $a)\ \frac{\cos2 \alpha }{1+\sin2 \alpha }=\frac{\cos^2 \alpha -\sin^2 \alpha }{\sin^2 \alpha +\cos^2 \alpha +2\sin \alpha \cos \alpha }=\frac{(\cos \alpha -\sin \alpha )(\cos \alpha +\sin \alpha )}{(\cos \alpha +\sin \alpha )^2}=\\\\=\frac{\cos \alpha -\sin \alpha }{\cos \alpha +\sin \alpha }=\frac{\cos \alpha (1-\frac{\sin \alpha }{\cos \alpha })}{\cos \alpha (1+\frac{\sin \alpha }{\cos \alpha })}=\frac{1-tg \alpha }{1+tg \alpha };$

$b)\ 2\cos(45^\circ+ \alpha )\bullet\cos(45^\circ- \alpha 
)=\\\\2\bullet\frac{1}{2}(\cos((45^\circ+ \alpha) -(45^\circ- \alpha 
))+\cos(45^\circ+ \alpha +45^\circ- \alpha ))=\\\\\cos2 \alpha 
+\cos90^\circ=\cos2 \alpha +0=\cos2 \alpha .$

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2...

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