Writing this expression as a single trig function? $2\cos^2\left(\frac{\pi}6\right) - 1$

Question

How can I write the following as a single trig function?

$$2\cos^2\left(\frac{\pi}6\right) - 1$$

To attempt this, I looked at all the identities (sum & difference, double angle identities, and Pythagorean identities)** and the identity that seemed fitting was the Pythagorean identity specifically $\sin^2\theta + \cos^2\theta = 1$.

I used this identity to isolate $\cos^2\theta$, but then I found myself using it over and over again; it was alternating between subbing in sine squared and cosine squared. (I didn't bother to post the work, but I can if it's needed.)

How should I get the textbook answer of $\cos \frac{\pi}3$?

A hint to an approach I might've missed would be grateful! :)

**I've been merely taught the three identities listed. Unfortunately, I can't identify any other methods mentioned in the answers.

Answer 1

The key fact to notice here is that the angles we're dealing with differ by a factor of two, i.e.

$$2 \cdot \frac \pi 6 = \frac \pi 3 \iff \frac 1 2 \cdot \frac \pi 3 = \frac \pi 6$$

This suggests immediately the use of the double- or half-angle formulas.

---

Recall the half-angle identity for cosine:

$$\cos \frac \theta 2 = \pm \sqrt{\frac{1 + \cos\theta}{2}}$$

Solving for $\cos\theta$,

$$\cos \theta = -1 + 2 \cos^2 \frac \theta 2$$

Take $\theta = \pi/3$ for your answer.

---

Alternatively, use (one of) the double-angle identity for cosine:

$$\cos 2\theta = 2 \cos^2 \theta - 1$$

With $\theta = \pi/6$, your starting expression is on the right.

Answer 2

$\cos (a+b)=\cos \,a \cos \, b-\sin \,a \sin \,b$. Put $a=b$ to get $\cos(2a)=\cos^{2}(a)-\sin^{2}(a)=2\cos^{2}(a)-1$. Put $a= \frac {\pi} 6$

Answer 3

I think, it's better to remember that $$2\cos^2\alpha-1=\cos2\alpha.$$