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

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Answer 1 · 2016-02-09T06:13:14

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

As a decimal: #.06699#

#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