How do you prove #cos36*cos72=1/4#?

1 Answer
Apr 6, 2016

Assuming you do not already know the values of the two cosines, you can find both the difference and the product using the double-angle and sum-product relations for cosine. See the explanation.

Explanation:

First you should prove that

#\cos(36°)-\cos(72°)=1/2#

To do this let #x# be the above difference and take its square:

#x^2=\cos^2(36°)-2\cos(36°)\cos(72°)+\cos^2(72°)#

Apply trigonometric identities you should know. From the double angle formula for cosine:

#\cos^2(36°)+\cos^2(72°)=((1+\cos(72°))/2)+((1+\cos(144°))/2)#

#=1+((\cos(72°)-\cos(36^o))/2)#

#=1-x/2#

Then from the sum-product relation for cosines:

#2\cos(36°)\cos(72°)=\cos(108°)+\cos(36°) =\cos(36°)-\cos(72°)=x#

Put these results together to get:

#x^2=1-(x/2)-x=1-(3/2)x#

Thus

#2x^2-3x-2=0#

This has two roots #x=1/2# and #x=-2#. We know that #x# must be positive so:

#x=\cos(36°)-\cos(72°)=1/2#

Now the product is immediately known because we already calculated:

#2\cos(36°)\cos(72°)=x#

So #\cos(36°)\cos(72°)=1/4#.