How do you evaluate #cos 15#?

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Answer 1 · 2015-11-06T23:06:46

#cos15=(sqrt(sqrt(3)+2))/2#

To find this value we can use the fact, that we know #cos30#
(#cos30=sqrt(3)/2#)

#cos30=cos2*15=cos^2 15-sin^2 15#

On the other hand #sin^2 15+cos^2 15=1#, so we can make a system of equations:

To simplify the notation I will use #s# for #sin15# and #c# for #cos15#

#{(c^2+s^2=1),(c^2-s^2=sqrt(3)/2) :}#

If we add both sides of the equations we get:

#2c^2=(2+sqrt(3))/2#

#c^2=(2+sqrt(3))/4#

And finally #c=(sqrt(2+sqrt(3)))/2#