5분만 ㄱㄷ

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2)는 11/12

1)  3/2*pi<x<2pi이고, cosx=4/5이므로 sinx=-3/5
따라서 sin(1/2x-3/2pi)=sin(1/2x)cos(3/2pi)-cos(1/2x)sin(3/2pi)
sin(3/2pi)=-1, cos(3/2pi)=0, cos(1/2x)=sqrt((1-cosx)/2)=1/sqrt10이므로
답은 1/10인듯허네요
2) (3+2/3)/4 = 11/12

1 풀고있었는데 ....

어 틀렸다
마지막에서 cos(1/2x)=sqrt((1+cosx)/2)=sqrt(9/10)=3/sqrt10해서
3/sqrt10이 답인듯 허네요

헐..사쿠라기 변호사가 뭔지 아는사람이 등장한것인가!

제..제길 cosx가 3/5라니
그럼... 어쨋든
cos(1/2x)^2 = (1+cosx)/2 이공식은 알거고...
그래서 숫자대입해보면 cos(1/2x)=sqrt((1+cosx)/2)=-2/sqrt5  ( 3/4pi<1/2x<pi이므로negative)
위에 식(sin(1/2x-3/2pi)=sin(1/2x)cos(3/2pi)-cos(1/2x)sin(3/2pi))에 다 대입해보면..
답이 -2/sqrt5

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