Question

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

Answer 1

`((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`