Saturday, December 24, 2011

`(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)`

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