How do you evaluate #cos((7pi)/6+pi/4)#?

1 Answer
Apr 7, 2016

#((sqrt2)(1 - sqrt3))/4#

Explanation:

Apply trig identity: cos (a + b) = cos a.cos b - sin a.sin b
#cos ((7pi)/6 + (pi)/4) = cos ((7pi)/6).cos (pi/4) - sin (pi/4).sin ((7pi)/6)#
Trig table gives and trig unit circle give-->
#cos (pi/4) = sqrt2/2#; #sin (pi/4) = sqrt2/2#
#cos ((7pi)/6) = - cos (pi/6) = - sqrt3/2#
#sin ((7pi)/6) = - sin (pi/6) = -1/2#
Therefor,

#cos ((7pi)/6 + pi/4) = (-sqrt3/2)(sqrt2/2) - (sqrt2/2)(-1/2) = #
#= ((sqrt2)(1 - sqrt3))/4#