How can Sine, Cosine and Tangent be found?

#(5pi)/12#

1 Answer
George C.
Mar 9, 2018

#sin((5pi)/12) = 1/4(sqrt(6)+sqrt(2))#

#cos((5pi)/12) = 1/4(sqrt(6)-sqrt(2))#

#tan((5pi)/12) = 2+sqrt(3)#

Explanation:

Using the sum of angles formula for #sin# we find:

#sin((5pi)/12) = sin(pi/4+pi/6)#

#color(white)(sin((5pi)/12)) = sin(pi/4)cos(pi/6)+sin(pi/6)cos(pi/4)#

#color(white)(sin((5pi)/12)) = sqrt(2)/2 sqrt(3)/2+1/2 sqrt(2)/2#

#color(white)(sin((5pi)/12)) = 1/4(sqrt(6)+sqrt(2))#

Using the sum of angles formula for #cos# we find:

#cos((5pi)/12) = cos(pi/4+pi/6)#

#color(white)(cos((5pi)/12)) = cos(pi/4)cos(pi/6) - sin(pi/4)sin(pi/6)#

#color(white)(cos((5pi)/12)) = sqrt(2)/2 sqrt(3)/2 - sqrt(2)/2 1/2#

#color(white)(cos((5pi)/12)) = 1/4(sqrt(6)-sqrt(2))#

Then:

#tan((5pi)/12) = sin((5pi)/12)/cos((5pi)/12)#

#color(white)(tan((5pi)/12)) = (1/4(sqrt(6)+sqrt(2)))/(1/4(sqrt(6)-sqrt(2)))#

#color(white)(tan((5pi)/12)) = (sqrt(6)+sqrt(2))^2/((sqrt(6)-sqrt(2))(sqrt(6)+sqrt(2)))#

#color(white)(tan((5pi)/12)) = (6+4sqrt(3)+2)/(6-2)#

#color(white)(tan((5pi)/12)) = 2+sqrt(3)#

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