How do you find the value of #cos(pi/3 - pi/6)#?
2 Answers
Apr 11, 2016
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.