Question

1) tgα= 1/3, найти sin(2α-7π/4)2) sinα = -3/5, π<a<3π/2, cosβ = 15/17, 3π/2<β<2, найти sin2α , cos2β , sin(α+β), cos(α-β).

Answer 1

1)tga=1/3
cos²a=1:(1+tg²a)=1:(1+1/9)=1:10/9=9/10
cosa=3/√10
sina=√(1-cos²a)=√(1-9/10)=1/√10
sin2a=2sinacosa=2*1/√10*3/√10=3/5
cos2a=cos²a-sin²a=9/10-1/10=4/5
sin(2a-7π/4)=sin(2a+π/4)=sin2acosπ/4+cos2asinπ/4=3/5*√2/2+4/5*√2/2=
=√2/2(3/5+4/5)=√2/2*7/5=7√2/10

2)sin2a=2sinacosa=2*(-3/5)*(-√(1-9/25))=-6/5*(-4/5)=24/25
cos2b=2cos²b-1=2*225/289-1=(450-289)/289=161/289
sin(a+b)=sinacosb+cosasinb=-3/5*15/17+(-√(1-9/25))*(-√(1-225/289))=
=-45/85+(-4/5)*(-8/17)=-45/85+32/85=-13/85
cos(a-b)=cosacosb+sinasinb=-4/5*15/17+(-3/5)*(-8/17)=-60/85+24/85=-36/85