`(13pi)/12` Find the exact values of the sine, cosine, and tangent of the angle.

`sin((13pi)/12)=sin(pi/2+pi/3+pi/4)`

As we know that `sin(pi/2+theta)=cos(theta)`

`:.sin((13pi)/12)=cos(pi/3+pi/4)`

Now use the identity `cos(x+y)=cos(x)cos(y)-sin(x)sin(y)`

`sin((13pi)/12)=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)`

`sin((13pi)/12)=((1/2)(1/sqrt(2))-((sqrt(3)/2)(1/sqrt(2)))`

`sin((13pi)/12)=(1-sqrt(3))/(2sqrt(2))`

`sin((13pi)/12)=(sqrt(2)-sqrt(6))/4`

`cos((13pi)/12)=cos(pi/2+pi/3+pi/4)` 

We know that `cos(pi/2+theta)=-sin(theta)`

`:.cos((13pi)/12)=-sin(pi/3+pi/4)`

using identity `sin(x+y)=sin(x)cos(y)+cos(x)sin(y)`

`cos((13pi)/12)=-(sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))`

`=-(sqrt(3)/2*1/sqrt(2)+1/2*1/sqrt(2))`

`=-(sqrt(3)+1)/(2sqrt(2))`

`=-(sqrt(6)+sqrt(2))/4` 

`tan((13pi)/12)=sin((13pi)/12)/cos((13pi)/12)`

plug in the values evaluated above,

`tan((13pi)/12)=((sqrt(2)-sqrt(6))/4)/(-(sqrt(6)+sqrt(2))/4)`

`=(sqrt(2)-sqrt(6))/(-(sqrt(6)+sqrt(2)))`

rationalizing the denominator,

`=((sqrt(2)-sqrt(6))(sqrt(6)-sqrt(2)))/(-(6-2))`  

`=(sqrt(12)-2-6+sqrt(12))/(-4)`

`=(2sqrt(12)-8)/(-4)`

`=(2*2sqrt(3)-8)/(-4)`

`=2-sqrt(3)`