`(7pi)/12 = pi/3 + pi/4` Find the exact values of the sine, cosine, and tangent of the angle.

Expert Answers
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You need to evaluate the sine of `(7pi)/12` , using the formula `sin(a+b) = sin a*cos b + sin b*cos a` such that:

`sin ((7pi)/12)= sin(pi/3 + pi/4) = sin (pi/3)*cos (pi/4) + sin (pi/4)*cos (pi/3)`

`sin ((7pi)/12)=(sqrt3)/2*(sqrt2)/2 + (sqrt2)/2*1/2`

`sin ((7pi)/12) = (sqrt2)/2*(sqrt3 + 1)/2`

You need to evaluate the cosine of `(7pi)/12` , using the formula `cos(a+b) = cos a*cos b - sin b*sin a` such that:

`cos ((7pi)/12) = cos (pi/3 + pi/4) = cos (pi/3)*cos (pi/4)- sin ( pi/4)*sin (pi/3)`

`cos ((7pi)/12)= 1/2*(sqrt2)/2 - (sqrt2)/2*(sqrt3)/2`

`cos((7pi)/12) = (sqrt2)/2*(1 - sqrt3)/2`

You need to evaluate the tangent of `(7pi)/12` , such that:

`tan ((7pi)/12) = (sin((7pi)/12))/(cos ((7pi)/12))`

`tan ((7pi)/12) = ((sqrt2)/2*(sqrt3 + 1)/2)/((sqrt2)/2*(1 - sqrt3)/2)`

`tan((7pi)/12) = (sqrt3 + 1)/(1 - sqrt3)`

`tan((7pi)/12) = ((sqrt3 + 1)*(1 + sqrt3))/(1 - 3)`

`tan((7pi)/12)) = -((sqrt3 + 1)^2)/2`

Hence, evaluating the sine, cosine and tangent of `tan(7pi)/12` , yields `sin((7pi)/12 ) = (sqrt2)/2*(sqrt3 + 1)/2, cos ( (7pi)/12 ) = (sqrt2)/2*(1 - sqrt3)/2, tan (7pi)/12 = -((sqrt3 + 1)^2)/2.`