How do you find the value of #cos(pi/3 - pi/6)#?

2 Answers
Apr 11, 2016

#1/2 sqrt3#

Explanation:

Using the appropriate #color(blue)" Addition formula " #

#color(red)(|bar(ul(color(white)(a/a)color(black)(cos(A +- B)= cosAcosB ∓ sinAsinB )color(white)(a/a)|)))#

#rArr cos(pi/3 - pi/6) = cos(pi/3)cos(pi/6) + sin(pi/3)sin(pi/6)#

Now using the #color(blue)" exact value triangle "#

with angles of # pi/6 , pi/3 , pi/2" and sides " 1 , 2 , sqrt3 " we obtain "#

#cos(pi/3) = 1/2 , cos(pi/6) = (sqrt3)/2 #

and # sin(pi/3) = (sqrt3)/2 , sin(pi/6) = 1/2 #

substituting these values into the right side of the expansion.

#= [1/2xx(sqrt3)/2] +[ (sqrt3)/2xx 1/2] = 1/4 sqrt3 + 1/4 sqrt3 = 1/2 sqrt3#

Apr 11, 2016

I would do the subtraction first.

Explanation:

#cos(pi/3-pi/6) = cos((2pi)/6-pi/6) = cos(pi/6) = sqrt3/2#