Опростяване на тригонометричен израз

[rawrzxz]

[tex]1)cotg \frac{\alpha}{ 2} = - \frac{4}{3 } ; \alpha \in (270^\circ ;360^\circ )

Izchisli : A=sin\alpha +cos\alpha ...........2)


Oprosti \frac{sin2\alpha }{ 1+cos2\alpha } .\frac{cos\alpha }{1+cos\alpha } = ?[/tex]

[Xixibg]

[tex]1)cotg \frac{\alpha}{ 2} = - \frac{4}{3 } ; \alpha \in (270^\circ ;360^\circ )

Izchisli : A=sin\alpha +cos\alpha ...........2)


Oprosti \frac{sin2\alpha }{ 1+cos2\alpha } .\frac{cos\alpha }{1+cos\alpha } = ?[/tex]

[tex]\alpha \in (270^\circ ;360^\circ ) ; =>\frac{\alpha }{2}\in (135^\circ ;180^\circ )[/tex]
[tex]=>sin{\frac{\alpha }{2}}=\frac{3}{5} ; cos {\frac{\alpha }{2}}=-\frac{4}{5}[/tex]
[tex]=>sin \alpha =2.sin{\frac{\alpha }{2}}.cos {\frac{\alpha }{2}}=-\frac{24}{25}[/tex]
[tex]=>cos \alpha =cos^2 {\frac{\alpha }{2}}-sin^2{\frac{\alpha }{2}}=\frac{7}{25}[/tex]
[tex]=>A=sin\alpha +cos\alpha=-\frac{17}{25}[/tex]

[tex]\frac{sin2\alpha }{ 1+cos2\alpha } .\frac{cos\alpha }{1+cos\alpha } =\frac{2sin \alpha.cos \alpha }{ 2cos^2\alpha } .\frac{cos\alpha }{1+cos\alpha } =\frac{sin\alpha }{1+cos\alpha }=\frac{2sin {\frac{\alpha }{2}}.cos {\frac{\alpha }{2}}}{2cos^2 {\frac{\alpha }{2}}}=tg {\frac{\alpha }{2}}[/tex]

[Гост]

Едно голямо благодаря !