# Find the exact value of the sine, cosine, and tangent of 5π/12 by setting α = π/4 and β = π/6 in one of the formulas previously derived.Show complete solution and explain the answer

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You may write `5pi/12 = pi/4 + pi/6= 3pi/12 + 2pi/12`

You may use the formula sin(x+y) = sin x*cos y + sin y*cos x such that:

`sin(pi/4 + pi/6) = sin pi/4*cos pi/6 + sin pi/6* cos pi/4`

`sin(pi/4 + pi/6) = (sqrt2/2)*(sqrt3/2) + (1/2)*(sqrt2/2)`

`sin(pi/4 + pi/6) = (sqrt6 + sqrt2)/4`

You may use the formula cos(x+y) = cos x*cos y - sin y*sin x such that:

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

`cos(pi/4 + pi/6) = (sqrt6- sqrt2)/4`

You may use the formula `tan alpha = sin alpha/cos alpha` such that:

`tan 5pi/12 = sin(pi/4 + pi/6)/cos(pi/4 + pi/6)`

`tan 5pi/12 = ((sqrt6 + sqrt2)/4)/((sqrt6- sqrt2)/4)`

`tan 5pi/12 = (sqrt6 + sqrt2)/(sqrt6- sqrt2)`

`tan 5pi/12 = (sqrt6 + sqrt2)^2/(6- 2)`

`tan 5pi/12 = (8 + 4sqrt3)/4 =gt tan 5pi/12 = (2 + sqrt3)`

**Hence, evaluating the values of functions sine, cosine and tangent at`5pi/12` yields: `sin(5pi/12) = (sqrt6 + sqrt2)/4 ; cos(5pi/12) =(sqrt6- sqrt2)/4;tan 5pi/12 = (2 + sqrt3).` **