Question

L3 Calculus:

Simplify the trigonometric expression:

3
-----------------------
5cos^3(pi/3)

Help would be greatly appreciated, step by step. Thanks xx​

Answer 1

`\cos\frac\pi3=\frac12`

so you have

`\frac3{5\cos^3\frac\pi3}=\frac3{5\left(\frac12\right)^3}=\frac3{\frac58}=\frac{24}5`

###

If you don't remember the value of
`\frac\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=\frac\pi3`. Then

`\cos\pi=4\cos^3\frac\pi3-3\cos\frac\pi3`

Let
`y=\cos\frac\pi3`. Then

`-1=4y^3-3y`

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

The rational root theorem suggests some possible roots are

`\pm\frac14,\pm\frac12,\pm1`

and checking all of these, we find that
`y=\frac12` is among the solution set. In fact,

`4y^3-3y+1=(y+1)\left(y-\frac12\right)^2=0\implies y=-1\text{ or }y=\frac12`

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

`\cos\frac\pi3=\frac12`