Question

L3 Calculus:

Answer 1

$\cos\dfrac\pi3=\dfrac12$

so you have

$\dfrac3{5\cos^3\frac\pi3}=\dfrac3{5\left(\frac12\right)^3}=\dfrac3{\frac58}=\dfrac{24}5$

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If you don't remember the value of $\dfrac\pi3$ off the top of your head, it's possible to derive it with some identities and knowing that $\cos\pi=-1$.

Consider the expression $\cos3x$. With the angle sum identity, we have

$\cos3x=\cos x\cos2x-\sin x\sin2x$

and the double angle identities give

$\cos3x=\cos x(\cos^2x-\sin^2x)-2\sin^2x\cos x$

Write everything in terms of cosine:

$\cos3x=\cos x(2\cos^2x-1)-2(1-\cos^2x)\cos x$

$\cos3x=4\cos^3x-3\cos x$

Now let $x=\dfrac\pi3$. Then

$\cos\pi=4\cos^3\dfrac\pi3-3\cos\dfrac\pi3$

Let $y=\cos\dfrac\pi3$. Then

$-1=4y^3-3y$

$4y^3-3y+1=0$

The rational root theorem suggests some possible roots are

$\pm\dfrac14,\pm\dfrac12,\pm1$

and checking all of these, we find that $y=\dfrac12$ is among the solution set. In fact,

$4y^3-3y+1=(y+1)\left(y-\dfrac12\right)^2=0\implies y=-1\text{ or }y=\dfrac12$

We have $\cos x=-1$ only for odd multiples of $\pi$, so it follows that

$\cos\dfrac\pi3=\dfrac12$