How do you prove cos36*cos72=1/4cos36cos72=14?

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.