# Using the expansions of sin(A-B) and cos(A-B),show that Sin `(pi/12)= (sqrt6-sqrt2)/4` and `cos (pi/12) = (sqrt6+sqrt2)/4`

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### 1 Answer

`sin(A-B) = sinAcosB-cosAsinB`

`cos(A-B) = cosAcosB+sinAsinB`

`pi/12 = pi/3-pi/4`

Let `A = pi/3 ` and` B = pi/4`

`sin(A-B) = sinAcosB-cosAsinB`

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

`sin(pi/12) = sqrt3/2xx1/sqrt2-1/2xx1/sqrt2`

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

`sin(pi/12) = (sqrt3-1)/(2sqrt2)xxsqrt2/sqrt2`

`sin(pi/12) = (sqrt6-sqrt2)/(2xx2)`

`sin(pi/12) = (sqrt6-sqrt2)/4`

`cos(A-B) = cosAcosB+sinAsinB`

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

`cos(pi/12) = 1/2xx1/sqrt2+sqrt3/2xx1/sqrt2`

`cos(pi/12) =(1+sqrt3)/(2sqrt2)` using same calculations as before.

`cos(pi/12) =(sqrt2+sqrt6)/4`

*So the required answers are proved as above.*

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