Из рисунка видно:
<AOB =<AOX+<BOX=45°+ α ; tqα =3/1=3;
tq(<AOB)= tq(45+α) = (tq45°+tqα)/(1-tq45°*tqα)= (1+tqα)/(1-tqα) =(1+3)/(1-3) = - 2 ;
1+tq²α =1/cos²α ;
cosα = (+/-) 1/sqrt(1+tq²α) =(+/-)sqrt(1+(-2)²) =(+/-)√5; т.к. tqα < 0 ==> 90°< α <180° ,
где cosα < 0 ==> cosα = -1/√5 , следовательно
3√5cosα = 3√5*(-1/√5) = - 3 ;
cos(<AOB)=cos(45 +α)= √2/2(cosα - sinα) =√2/2(1/√10 - 3/√10)=
= -1/√5.
(х+3) (х^2-3х+9)=х³+3³=х³+27
<span>(2а-3b) (4а^2+6ab+9b^2)=(2а)</span>³-(3b)³=8a³-27b³
![f'(x) = \frac{-2}{ sin^{2} x}](https://tex.z-dn.net/?f=f%27%28x%29+%3D++%5Cfrac%7B-2%7D%7B+sin%5E%7B2%7D+x%7D+)
f"(x) = - 2 / (sin^(2) x)
tg x = 5
x = arctg 5 + Пn (ПИ умножить на n )
tg x = 1
x = arctg 1 + Пn