Question

How do you evaluate `(sin (π/ 6) cos (π /4) − sin (π/ 4) cos (π/ 6) )^2`?

Answer 1

`((2 - sqrt(3))/4)`

As a decimal: `.06699`

Explanation:

`sin(pi/6) = 1/2`
`cos(pi/6) = sqrt(3)/2`
(http://mathworld.wolfram.com/TrigonometryAnglesPi6.html)

`sin(pi/4) = cos(pi/4) = sqrt(2)/2`

Now let's plug in:

`(1/2*sqrt(2)/2 - sqrt(2)/2*sqrt(3)/2)^2`

multiply the two terms:
`(sqrt(2)/4 - sqrt(6)/4)^2`

Consolidate the numerators:
`((sqrt(2) - sqrt(6))/4)^2`

Square the numerator and denominator:
`((2 - 2sqrt(12) + 6)/16)`

Gather like terms (2 and 6):
`((8 - 2sqrt(12))/16)`

Factor `sqrt(12) -> sqrt(4) * sqrt(3)`

`((8 - 2*sqrt(4)*sqrt(3))/16)`

`sqrt(4) = 2`

`((8 - 2*2*sqrt(3))/16)`

simplify:

`((8 - 4*sqrt(3))/16)`

Divide each term by 4:

`((2 - sqrt(3))/4)`

As a decimal: .06699