![\cos\dfrac\pi3=\dfrac12](/tpl/images/0476/5418/e023b.png)
so you have
![\dfrac3{5\cos^3\frac\pi3}=\dfrac3{5\left(\frac12\right)^3}=\dfrac3{\frac58}=\dfrac{24}5](/tpl/images/0476/5418/9433a.png)
###
If you don't remember the value of
off the top of your head, it's possible to derive it with some identities and knowing that
.
Consider the expression
. With the angle sum identity, we have
![\cos3x=\cos x\cos2x-\sin x\sin2x](/tpl/images/0476/5418/c2f27.png)
and the double angle identities give
![\cos3x=\cos x(\cos^2x-\sin^2x)-2\sin^2x\cos x](/tpl/images/0476/5418/643d2.png)
Write everything in terms of cosine:
![\cos3x=\cos x(2\cos^2x-1)-2(1-\cos^2x)\cos x](/tpl/images/0476/5418/318fd.png)
![\cos3x=4\cos^3x-3\cos x](/tpl/images/0476/5418/4374a.png)
Now let
. Then
![\cos\pi=4\cos^3\dfrac\pi3-3\cos\dfrac\pi3](/tpl/images/0476/5418/5b37e.png)
Let
. Then
![-1=4y^3-3y](/tpl/images/0476/5418/55c97.png)
![4y^3-3y+1=0](/tpl/images/0476/5418/06e2a.png)
The rational root theorem suggests some possible roots are
![\pm\dfrac14,\pm\dfrac12,\pm1](/tpl/images/0476/5418/7ad68.png)
and checking all of these, we find that
is among the solution set. In fact,
![4y^3-3y+1=(y+1)\left(y-\dfrac12\right)^2=0\implies y=-1\text{ or }y=\dfrac12](/tpl/images/0476/5418/a0031.png)
We have
only for odd multiples of
, so it follows that
![\cos\dfrac\pi3=\dfrac12](/tpl/images/0476/5418/e023b.png)