Tożsamość trygonometryczna

[technikmaliniak]

Ta nieszczęsna tożsamość \(\displaystyle{ \frac{\sin x + \cos (2y-x) }{\cos x - \sin (2y-x)}=\frac{1+\sin 2y}{\cos 2y}}\) Z góry dziękuje, za każda pomoc

[anna_]

\(\displaystyle{ \frac{\sin x + \cos (2y-x) }{\cos x - \sin (2y-x)}=\frac{\sin x + \sin(90^o-2y+x) }{\cos x - \cos (90^o-2y+x)}=\frac{2\sin \frac{x+90^o-2y+x}{2}cos \frac{x-90^o+2y-x}{2} }{-2\sin \frac{x+90^o-2y+x}{2}sin \frac{x-90^o+2y-x}{2} }=-\frac{cos \frac{-90^o+2y}{2} }{sin \frac{-90^o+2y}{2} }=-ctg\frac{-90^o+2y}{2}=ctg\frac{90^o-2y}{2}= \frac{1+cos(90^o-2y)}{sin(90^o-2y)}=\frac{1+sin2y}{cos2y}}\)