Малко тригонометрични задачки

[Spider Iovkov]

Да се докаже, че ако [tex]\alpha+\beta+\gamma=180^\circ[/tex], то:
а) [tex]sin\alpha+sin\beta+sin\gamma=4cos{\frac{\alpha}{2}}cos{\frac{\beta}{2}}cos{\frac{\gamma}{2}}[/tex];
б) [tex]cos\alpha+cos\beta+cos\gamma=1+4sin{\frac{\alpha}{2}}sin{\frac{\beta}{2}}sin{\frac{\gamma}{2}}[/tex];
в) [tex]sin2k\alpha+sin2k\beta+sin2k\gamma=4(-1)^{k+1}sink\alpha sink\beta sin k\gamma[/tex].

[Реклама]

[gvateva]

а) от даденото, че [tex] \alpha +\beta +\gamma =180^0 [/tex] следва, че:
[tex] \alpha +\beta =180^0-\gamma [/tex] и [tex] \gamma =180^0-(\alpha +\beta ) [/tex]
[tex] A=(sin\alpha +sin\beta) +sin\gamma [/tex]
[tex] A=2sin(\frac{\alpha +\beta }{ 2}).cos(\frac{\alpha-\beta }{2 }) +sin [180^0-(\alpha +\beta )] [/tex]
[tex] A=2sin(\frac{\alpha +\beta }{ 2}).cos(\frac{\alpha-\beta }{2 }) +sin(\alpha +\beta ) [/tex]
[tex] A=2sin(\frac{\alpha +\beta }{ 2}).cos(\frac{\alpha-\beta }{2 })+2sin(\frac{\alpha +\beta }{ 2} ).cos(\frac{\alpha +\beta }{ 2} ) [/tex]
[tex] A= 2sin(\frac{\alpha +\beta }{ 2}).[cos(\frac{\alpha-\beta }{ 2} )+cos(\frac{\alpha +\beta }{2 } )] [/tex]
[tex] A=2sin(\frac{180^0-\gamma }{2 } ).2.cos(\frac{\alpha }{2 }).cos(\frac{-\beta }{2 }) [/tex]
[tex] A=4.cos(\frac{\gamma }{2 } ).cos(\frac{\alpha }{2 }).cos(\frac{\beta }{2 }) [/tex]

[gvateva]

б)
[tex] B=(cos\alpha +cos\beta )+cos\gamma [/tex]
[tex] B= 2cos(\frac{\alpha+\beta }{ 2}).cos(\frac{\alpha -\beta }{2 } )+cos [180^0-(\alpha +\beta )] [/tex]
[tex] B= 2cos(\frac{\alpha+\beta }{ 2}).cos(\frac{\alpha -\beta }{2 } )-cos(\alpha+\beta )[/tex]
[tex] B= 2cos(\frac{\alpha+\beta }{ 2}).cos(\frac{\alpha -\beta }{2 } ) - [2cos^2(\frac{\alpha+\beta }{ 2} )-1] [/tex]
[tex] B= 1+ 2cos(\frac{\alpha+\beta }{ 2}).[cos(\frac{\alpha-\beta }{ 2})-cos(\frac{\alpha+\beta }{ 2})] [/tex]
[tex] B=1+ 2cos(\frac{\alpha+\beta }{ 2}).2sin(\frac{\alpha }{ 2}).sin(\frac{\beta }{ 2} ) [/tex]
[tex] B= 1+2.cos(\frac{180^0-\gamma }{ 2} ).2sin(\frac{\alpha }{ 2}).sin(\frac{\beta }{ 2} ) [/tex]
[tex] B=1+4.sin(\frac{\gamma }{2 } ).sin(\frac{\alpha }{ 2}).sin(\frac{\beta }{ 2} ) [/tex]

[Анна]

Окръжността k(O,r)е външновписана за триъгълника ABC и се допира до BC.Да се намерят:r,AO,BO,CO,ако Pabc=2p,а ъглите му:α,β,γ.
Моля ви за някакво логично решение,защото аз ненамирам такова.Благодаря.

[ганка симеонова]

[tex]AP=AQ=p; \Delta AOQ=>r=ptg{\frac{\alpha }{ 2}}; AO=\frac{AQ}{cos{\frac{\alpha }{ 2}} }=\frac{p}{cos{\frac{\alpha }{ 2}} } [/tex]
[tex]\Delta BQO=>BO=\frac{r}{cos{\frac{\beta }{2} } }; \Delta CPO=>CO= \frac{r}{cos{\frac{\gamma }{2} } }[/tex]