Dziwna tożsamośc trygonometryczna

[goodspeed]

\(\displaystyle{ \frac{sin2x}{1+cos2x} * \frac{cosx}{1+cosx} =\tan \frac{x}{2}}\)


Z góry dziękuje za pomoc

[dr_grucha]

\(\displaystyle{ \frac{\sin 2x}{1+\cos 2x} \frac{\cos x}{1 + \cos x} = \frac{2 \sin x \cos x}{1+2 \cos ^{2} x - 1} \frac{\cos x}{1 + 2 \cos ^{2} \frac{x}{2} -1 } = \frac{2 \sin x \cos x}{2 \cos ^{2} x} \frac{\cos x}{2 \cos ^{2} \frac{x}{2} } = \frac{\sin x}{2 \cos ^{2} x } = \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2 \cos ^{2} \frac{x}{2} } = \frac{\sin \frac{x}{2} }{\cos \frac{x}{2} }= \tan \frac{x}{2}}\)