Question #12d17

1 Answer
Jul 5, 2016

26) and 27)

Explanation:

26) We know that
cos(x-y)=Cos(x) Cos(y) + Sin(x) Sin(y)cos(xy)=cos(x)cos(y)+sin(x)sin(y)
and
cos^2x+sin^2x=1cos2x+sin2x=1
then substituting
cos(x-y) = (sqrt[1 - (3/5)^2]) (5/13) + (3/5) sqrt[1 - (5/13)^2]cos(xy)=1(35)2(513)+(35)1(513)2

27) cos(2x) = sin x = cos(pi/2-x)cos(2x)=sinx=cos(π2x)
so
2x=pi/2-x+2k pi2x=π2x+2kπ
solving for xx
x = pi/6+2/3k pix=π6+23kπ for k = 0,pm1,pm2,...