How do you use the sum or difference identities to find the exact value of #sin(pi/12)#?
1 Answer
Jun 19, 2018
Explanation:
#"using the "color(blue)"difference identity for sin"#
#•color(white)(x)sin(x-y)=sinxcosy-cosxsiny#
#"note that "pi/12=pi/4-pi/6#
#sin(pi/12)=sin(pi/4-pi/6)#
#=sin(pi/4)cos(pi/6)-cos(pi/4)sin(pi/6)#
#=(sqrt2/2xxsqrt3/2)-(sqrt2/2xx1/2)#
#=sqrt6/4-sqrt2/4=1/4(sqrt6-sqrt2)#
