Задачи по Математика 11 клас. Преговор

[itzko_690554]

1. Опростете израза:

[tex]\frac{sin450^\circ.cos(\alpha-90^\circ).cos392^\circ.cos(180^\circ+\alpha)}{cotg32^\circ.sin148^\circ.sin(270^\circ+\alpha)}[/tex]

2 Като знаете, че [tex]cotgx=- \frac{5}{12} , x \epsilon ({ \frac{\pi}{2}; \pi})[/tex] , пресметнете sin x и cos x

Докажете тъждеството

A [tex]= \frac{cos(\alpha-\beta)-cos(\alpha+\beta)}{sin(\alpha-\beta)-sin(\alpha+\beta)} = \frac{cotg2\alpha-cotg\alpha}{1+cotg\alpha.cotg2\alpha}[/tex]

и пресметнете числената стойност на А при [tex]\alpha = 30^\circ , \beta= 150^\circ[/tex]

[Nathi123]

1.зад. [tex]sin450^\circ = sin(360^\circ + 90^\circ )=sin90^\circ = 1 ; cos(\alpha - 90^\circ)=cos(90^\circ - \alpha )= sin\alpha ;[/tex]
[tex]cos392^\circ = cos(360^\circ +32^\circ) = cos32^\circ ;cos(180^\circ +\alpha) = -cos\alpha ; sin148^\circ = sin( 180^\circ - 32^\circ) = sin32^\circ[/tex];
[tex]sin( 270^\circ + \alpha ) = -cos\alpha[/tex]
[tex]\Rightarrow \frac{sin450^\circ.cos(90^\circ-\alpha).cos392^\circ.cos(180^\circ+\alpha)}{cotg32^\circ.sin148^\circ .sin(270^\circ +\alpha)} = \frac{sin\alpha cos32^\circ(-cos\alpha)}{cotg32^\circ sin32^\circ(-cos\alpha)} = sin\alpha[/tex].

[Nathi123]

2.зад. cotgx = - [tex]\frac{5}{12} ; x\in ( \frac{\pi}{2},\pi )\Rightarrow sinx >0 ; cosx <0 \Rightarrow cosx = - 5t ; sinx = 12t ; sin^{2}x + cos^{2}x = 1[/tex]
[tex]\Rightarrow 144t^{2} +25t^{2} =1 \Rightarrow 169t^{2}=1\Rightarrow t=\frac{1}{13} \Rightarrow sinx =\frac{12}{13} ; cosx = -\frac{5}{13}[/tex].

[Nathi123]

A=[tex]\frac{cos(\alpha - \beta) - cos ( \alpha+ \beta)}{sin(\alpha - \beta)- sin ( \alpha+ \beta)} = \frac{-2 sin\alpha.sin(-\beta)}{2 sin( -\beta)cos\alpha} = -tg\alpha[/tex].
[tex]B = \frac{cotg2\alpha - cotg\alpha}{1+cotg2\alpha.cotg\alpha} = \frac{1}{\frac{1+cotg2\alpha cotg\alpha}{cotg2\alpha - coyg\alpha}} = \frac{1}{cotg(\alpha - 2\alpha)} = \frac{1}{cotg(-\alpha)} = - tg\alpha \Rightarrow A = B[/tex].

[Nathi123]

При 3 задача A = B = - tg[tex]\alpha \Rightarrow A (30^\circ) = tg30^\circ =\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}[/tex].
От доказаното тъждество ,следва,че А и В не зависят от [tex]\beta[/tex].