Monday, December 6, 2010

`-pi/12 = pi/6 - pi/4` Find the exact values of the sine, cosine, and tangent of the angle.

You need to find the values of sine and cosine of `-pi/12` , using the formulas `sin(a-b) = sin a*cos b - sin b*cos a` and `cos(a-b) = cos a*cos b + sin a*sin b` , such that:


`sin(-pi/12) = sin(pi/6 - pi/4) = sin(pi/6)cos(pi/4) - sin(pi/4)cos(pi/6)`


`sin(-pi/12) = 1/2*sqrt2/2 - sqrt2/2*sqrt3/2`


`sin(-pi/12) =sqrt2/2*(1-sqrt3)/2`


`cos(-pi/12) = cos(pi/6 - pi/4) = cos(pi/6)cos(pi/4) + sin(pi/4)sin(pi/6)`


`cos(-pi/12) = sqrt3/2*sqrt2/2 + sqrt2/2*1/2`


`cos(-pi/12) = sqrt2/2*(1+sqrt3)/2`


You need to evaluate tangent function such that:


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


`tan(-pi/12) = (1-sqrt3)/(1+sqrt3)`


`tan(-pi/12) = -((1-sqrt3)^2)/2`


Hence, evaluating the values of the functions yields `sin(-pi/12) =sqrt2/2*(1-sqrt3)/2, cos(-pi/12) = sqrt2/2*(1+sqrt3)/2, tan(-pi/12) = -((1-sqrt3)^2)/2.`

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