Cos(a+b)+2sinasinb/2sinacosb-sin(a+b) (sina/sin2a-cosa/cos2a)*cosa-cos7a/sina - вопрос №375747822227 от svalevayulia10 15.12.2020 15:18

Question

Алгебра: Cos(a+b)+2sinasinb/2sinacosb-sin(a+b) (sina/sin2a-cosa/cos2a)*cosa-cos7a/sina sin(a+b)+sin(a-b)/sin(a+b)-sin(a-b) (sina-cosa)2-1+sin4a/cos2a+cos4a

svalevayulia10 · Answer

1)cos(a+b)+2sinasinb/2sinacosb-sin(a+b)=
(cosacosb-sinasinb+2sinasinb)/(2sinacosb-sinacosb-cosasinb)=
=(cosacosb+sinasinb)/(sinacosb-cosasinb)=cos(a-b)/sin(a-b)=ctg(a-b)
2)(sina/sin2a-cosa/cos2a)*cosa-cos7a/sina=
(sinacos2a-cosasin2a)/sin2acos2a* (-2sin3asin4a)/sina=
=2sin(a-2a)/sin4a* (-2sin3asin4a)/sina=-2sina*(-2sin3a)/sina=4sin3a
3)sin(a+b)+sin(a-b)/sin(a+b)-sin(a-b)=
=(sinacosb+cosasinb+sinacosb-cosasinb)/(sinacosb+cosasinb-sinacosb+
+cosasinb)=2sinacosb/2cosasinb=tga*ctgb
4)(sina-cosa)2-1+sin4a/cos2a+cos4a=
=(sin²a-2sinacosa+cos²a-sin²a-cos²a+sin4a)/2cos3acosa=
=(sin4a-sin2a)/2cos3acosa=2sinacos3a/2cos3acosa=sina/cosa=tga