Задача 35453 ...

hufflepuff

2019-04-06 16:10:28

Упростить :
2sin пи/12 × cos пи/12

cos2a+2sin²a

ctg²a(1-cos2a)²

1-cosa+sin2a/sin2a-sina

sova

2019-04-07 11:23:03

★

1.
формула синуса двойного угла:

2*sin(π/12)* cos(π/12)=sin2*(π/12)=sin(π/6)= [b]1/2[/b]

2.
формула косинуса двойного угла:

cos2a+2sin²a=сos^2a-sin^2a+2sin^2a=cos^2a+sin^2a= [b]1[/b]

3.

(1-сos2α)^2=(2sin^2α)^2=4sin^2α

ctg ^2α =cos^2 α /sin^2α

ctg^2α*(1-сos2α)^2=(cos^2 α /sin^2 α)* 4sin^2α =

=4sin^2 α *cos^2 α =(2sin α *cos α )^2= [b]sin^22 α [/b]

4.

1+cosα +sin2α =(1+cos α)+2*sin α *cos α =

=2sin^2( α /2)+2*2sin( α/2)*cos( α /2)*cos α=

=2sin( α /2) * ( sin( α/2)+ 2cos ( α /2) * cos α)



sin2α -sinα =2sinα *cosα-sinα = sinα*(2cos α -1)=

=2sin( α /2)*cos( α /2)*(2cos α -1)


(1+cosα +sin2α )/(sin2α - sinα) =

=2sin( α /2) * ( sin( α/2)+ 2cos ( α /2) * cos α)/(2sin( α /2)*cos( α /2)*
(2cos α -1))=

= ( sin( α/2)+ 2cos ( α /2) * cos α)/(cos( α /2)*(2cos α -1))=

=(tg( α /2)+2cos α )/(2cos α -1)