נושא : שימוש בזהויות s i n ( 180 − α ) 1 − c o s ( 180 − α ) − s i n ( 180 + α ) 1 + c o s ( 180 + α ) = 2 s i n ( α + 360 ) ? s i n ( 180 − α ) = s i n α s i n ( α + 360 ) = s i n α c o s ( 180 + α ) = − cos α s i n α 1 + c o s α − s i n ( 180 + α ) 1 + c o s ( 180 + α ) = 2 s i n α { s i n α = s i n ( 180 − α ) s i n α = s i n ( 180 − 180 − α ) s i n α = s i n ( − α ) sin ( − α ) = − sin α s i n α 1 + c o s α − − s i n ( α ) 1 + c o s ( 180 + α ) = 2 s i n α { cos ( 180 − α ) = − cos α c o s ( 180 − 180 + α ) = − cos α c o s α = − cos α s i n α 1 + c o s α − − s i n ( α ) 1 − c o s α = 2 s i n α s i n α ( 1 − cos α ) + s i n α ( 1 + c o s α ) ( 1 + cos α ) ( 1 − cos α ) = 2 s i n α s i n α − sin α ∗ cos α + s i n α + sin α ∗ cos α 1 − cos 2 α = 2 s i n α sin 2 α + cos 2 α = 1 2 s i n α s i n 2 α = 2 s i n α 2 s i n α = 2 s i n α {\displaystyle {\begin{aligned}&{\frac {sin(180-\alpha )}{1-cos(180-\alpha )}}-{\frac {sin(180+\alpha )}{1+cos(180+\alpha )}}={\frac {2}{sin(\alpha +360)}}?\\&\color {blue}sin(180-\alpha )=sin\alpha \\&\color {blue}sin(\alpha +360)=sin\alpha \\&\color {blue}cos(180+\alpha )=-\cos \alpha \\&{\frac {sin\alpha }{1+cos\alpha }}-{\frac {\color {green}sin(180+\alpha )}{1+cos(180+\alpha )}}={\frac {2}{sin\alpha }}\\&{\begin{cases}\color {blue}sin\alpha =sin(180-\alpha )\\sin\alpha ={\color {green}sin(180-180-\alpha )}\\sin\alpha =sin(-\alpha )\end{cases}}\\&\color {blue}\sin(-\alpha )=-\sin \alpha \\&{\frac {sin\alpha }{1+cos\alpha }}-{\frac {-sin(\alpha )}{1+{\color {green}cos(180+\alpha )}}}={\frac {2}{sin\alpha }}\\&{\begin{cases}\color {blue}\cos(180-\alpha )=-\cos \alpha \\cos(180-{\color {green}180+\alpha )}=-\cos \alpha \\cos\alpha =-\cos \alpha \end{cases}}\\&{\frac {sin\alpha }{1+cos\alpha }}-{\frac {-sin(\alpha )}{1-cos\alpha }}={\frac {2}{sin\alpha }}\\&{\frac {sin\alpha (1-\cos \alpha )+sin\alpha (1+cos\alpha )}{(1+\cos \alpha )(1-\cos \alpha )}}={\frac {2}{sin\alpha }}\\&{\frac {sin\alpha -\sin \alpha *\cos \alpha +sin\alpha +\sin \alpha *\cos \alpha }{1-\cos ^{2}\alpha }}={\frac {2}{sin\alpha }}\\&\color {blue}\sin ^{2}\alpha +\cos ^{2}\alpha =1\\&{\frac {2sin\alpha }{sin^{2}\alpha }}={\frac {2}{sin\alpha }}\\&{\frac {2}{sin\alpha }}={\frac {2}{sin\alpha }}\\\end{aligned}}}
