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

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Answer 1 · 2018-07-15T17:52:30

#(sqrt3+1)/2#

For problems like these, it helps to memorize the Unit Circle.

On the Unit Circle, the coordinates for #pi/6# (#30^@#) are

#(sqrt3/2,1/2)#

Where the #x#-coordinate is the cosine value and the #y#-coordinate is the sine value.

This means #cos(pi/6)=sqrt3/2# and #sin(pi/6)=1/2#. Adding them, we get

#(sqrt3+1)/2#

Hope this helps!