How do you find the exact values of the sine, cosine, and tangent of the angle #(5pi)/12#?

1 Answer
Mar 27, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)sin(x+y)=sinxcosy+cosxsiny#

#•color(white)(x)cos(x+y)=cosxcosy-sinxsiny#

#"note that "(5pi)/12=pi/4+pi/6#

#rArrsin((5pi)/12)=sin(pi/4+pi/6)#

#rArrsin(pi/4+pi/6)#

#=sin(pi/4)cos(pi/6)+cos(pi/4)sin(pi/6)#

#=(1/sqrt2xxsqrt3/2)+(1/sqrt2xx1/2)#

#=sqrt3/(2sqrt2)+1/(2sqrt2)#

#=(sqrt3+1)/(2sqrt2)xxsqrt2/sqrt2=1/4(sqrt6+sqrt2)larrcolor(red)"exact value"#

#cos((5pi)/12)=cos(pi/4+pi/6)#

#rArrcos(pi/4+pi/6)#

#=cos(pi/4)cos(pi/6)-sin(pi/4)sin(pi/6)#

#=(1/sqrt2xxsqrt3/2)-(1/sqrt2xx1/2)#

#=(sqrt3-1)/(2sqrt2)xxsqrt2/sqrt2#

#=1/4(sqrt6-sqrt2)larrcolor(red)"exact value"#

#tan((5pi)/12)=sin((5pi)/12)/cos((5pi)/12)#

#color(white)(xxxxxxx)=(sqrt6+sqrt2)/(sqrt6-sqrt2)xx(sqrt6+sqrt2)/(sqrt6+sqrt2)#

#color(white)(xxxxxxx)=(6+2sqrt12+2)/4#

#color(white)(xxxxxxx)=(8+4sqrt3)/4#

#color(white)(xxxxxxx)=2+sqrt3larrcolor(red)"exact value"#