Тригонометрия :)

[Гост]

1.Ако sinα = -3/5 и 180°< α < 270° пресметнете sin (α/2), cos(α/2), tg(α/2), cotg(α/2)

2. Представете като произведение 4cos^2 - 3

Хич, не ме бива в математика и тези две не можах да ги преборя, ако някой може, моля да ги реши

[Xixibg]

1.[tex]\begin{tabular}{|l}sin^2{\frac{\alpha }{2}}+cos^2{\frac{\alpha }{2}}=1\\2sin{\frac{\alpha }{2}}cos{\frac{\alpha }{2}}=sin\alpha =-\frac{3}{5}\end{tabular}[/tex]

[tex]=>\begin{tabular}{|l}(sin{\frac{\alpha }{2}}+cos{\frac{\alpha }{2}})^2=\frac{8}{5} ; =>sin{\frac{\alpha }{2}}+cos{\frac{\alpha }{2}}=\frac{2\sqrt{10}}{5}\\(sin{\frac{\alpha }{2}}-cos{\frac{\alpha }{2}})^2=\frac{2}{5} ; =>sin{\frac{\alpha }{2}}-cos{\frac{\alpha }{2}}=\frac{\sqrt{10}}{5}\end{tabular} ; =>sin{\frac{\alpha }{2}}=\frac{3\sqrt{10}}{10} ;cos{\frac{\alpha }{2}}=-\frac{\sqrt{10}}{10}[/tex]

[tex]tg\frac{\alpha }{2}=\frac{sin{\frac{\alpha }{2}}}{cos{\frac{\alpha }{2}}}=-3[/tex]

[tex]cotg \frac{\alpha }{2}=\frac{1}{tg{\frac{\alpha }{2}}}=-\frac{1}{3}[/tex]

2.[tex]4cos^2 \alpha -3=(2cos \alpha -\sqrt{3})(2cos \alpha+\sqrt{3})=4(cos \alpha -cos{\frac{\pi }{6}})(cos \alpha -cos{\frac{\pi }{6}})=-16sin{\frac{\alpha -60^\circ }{2}}sin{\frac{\alpha +60^\circ }{2}}cos{\frac{\alpha -60^\circ }{2}}cos{\frac{\alpha +60^\circ }{2}}[/tex]

[Гост]

zdraveite imam malko kolebanie za edna zadacha za izchislqvane
neka tgα=-2
sin,cos,cotg=?
Blagodarq