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Klenaya1Klenaya

25 апр. 2014 г., 22:12:05 (10 лет назад)

а) sin(5π/6-a)-1/2 cos a=
   =sin(5π/6)*cosa-cos(5π/6)*sina -1/2 cosa=
   = 1/2 cosa - √3/2 sina -1/2 cosa= -√3/2 sina

б) √3cos a- 2 cos(a-π/6)= 
    =√3 cosa -2cosa*cos (π/6)+sina*sin(π/6)=
    =√3 cosa -2cosa*√3/2 +sina*1/2 =√3cosa -√3cosa+1/2 sina=1/2 sina
   

в) √3/2 sin a+cos(a-5π/3)=√3/2 sina +cosa*cos(5π/3)+sina*sin(5π/3)=
    =√3/2 sina +cosa*1/2+sina*(-√3/2)=
    =√3/2 sina +1/2 cosa - √3/2 sina = 1/2 cosa

г) √2sin(a-π/4)-sin a =√2(sina*cos(π/4)-cosa*cos(π/4))-sina=
    =√2sina*cos(π/4)-√2cosa*cos(π/4)-sina=
    =√2 sina*√2/2 -√2 cosa*√2/2 -sina=
    =sina -cosa - sina=-cosa