Question

Помогите, пожалуйста!

$frac{1+sin2a}{cos(2a-2pi)ctg(a-frac{5}{4}pi)} +cos^2a$

Answer 1

1+sin2a/cos(2a-2pi)ctg(a-5/4pi)+cos2a=1+sin2a/cos(2pi-2a)(-ctg(5/4pi-a))+cos2a=1+sin2a/-cos2actg(4pi+pi/4-a)+cos2a=1+sin2a/cos2actg(pi+(pi/4-a))+cos2a=1+sin2a/cos2actg(pi/4-a)+cos2a=-cos2a+sin2a+2sinacosa/cos2a*cos(pi/4-a)/sin(pi/4-a)=+cos2a=-(cosa+sina)^2sin(pi/4-a)/cosacos(pi/4-a)+cos2a=-(cosa+sina)^2*$sqrt{ frac{2}{2} }$(cosa-sina)/(cosa+sina)(cosa-sina)*sqrt2/2(cosa+sina)+cos2a=-1+cos2a=(1-cos2a)=-sin2a.

Answer 2

1+sin(2a)=sin^2+cos^2+2*sin*cos=(sin+cos)^2
cos(2a-2pi)*ctg(a-5pi/4)=cos(2a)*ctg(a-pi/4)=-cos(2a)*(ctg(a)+1)/(ctg(a)-1)=-cos(2a)*(cos(a)+sin(a))/(cos(a)-sin(a))
(1+sin(2a))/(cos(2a-2pi)*ctg(a-5pi/4))+ cos^2(a) = -(sin+cos)^2 / (cos(2a)*(cos(a)+sin(a))/(cos(a)-sin(a))) + cos^2(a) =cos^2(a) -1