# Please show how calculate cos(`pi` /6)+cos(`pi` /3) with two methods?

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You need to evaluate the given summation using two methods, hence, I suggest you to begin with the easier method. Thus, you need to remember that `cos(pi/6) = sqrt3/2` and `cos(pi/3) = 1/2` , such that:

`cos(pi/6) + cos(pi/3) = sqrt3/2 + 1/2`

`cos(pi/6) + cos(pi/3) = (sqrt3 + 1)/2 ~~ 1.366`

You also evaluate the value of the given expression by converting the summation of cosine into a product, such that:

`cos(pi/6) + cos(pi/3) = 2*cos((pi/6 + pi/3)/2)*cos((pi/6 - pi/3)/2)`

`cos(pi/6) + cos(pi/3) = 2*cos(3pi/12)*cos(-pi/12)`

Since `cos(-a) = cos a` yields:

`cos(pi/6) + cos(pi/3) = 2*cos(pi/4)*cos(pi/12)`

You need to replace `sqrt2/2` for `cos(pi/4)` and you need to use the half angle identity to evaluate cos(pi/12), such that:

`cos(pi/6) + cos(pi/3) = 2*(sqrt2/2)*sqrt((1 + cos(pi/6))/2)`

Reducing duplicate factors yields:

`cos(pi/6) + cos(pi/3) = (sqrt2/2)*sqrt(2 + sqrt3)`

`cos(pi/6) + cos(pi/3) = (sqrt(2(2 + sqrt3)))/2`

`cos(pi/6) + cos(pi/3) = (sqrt(4+2sqrt3))/2 ~~ 1.366`

**Hence, evaluating the value of the given expression, either directly, by substitution, or converting the summation of cosines into a product, yields **`cos(pi/6) + cos(pi/3) ~~ 1.366.`

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