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

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.

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.

• I've been taught the identity as cos $2$θ = cos$^2$θ - sin$^2$θ, so do I just simply sub in $1$ if I want to use one of the double-angle identity for cosine? – Jenny B Feb 21 '19 at 8:43
• Not sure what you mean by that, but... Your identity and mine are actually equivalent. There are three double angle identities for cosine: $$\cos 2\theta = 2 \cos^2 \theta - 1 = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta$$ Each is handy in its own context. – Eevee Trainer Feb 21 '19 at 8:50
• ... and the outer two are derived from the one you know by using the Pythagorean identity. – NickD Feb 21 '19 at 15:51

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

• Did you get $1$ from sin$^2$($π \over 6$)? – Jenny B Feb 21 '19 at 8:39
• @JennyB I used the fact that $\sin^{2}(a)=1-\cos^{2}(a)$ so $\cos^{2}(a)-\sin^{2}(a)=2\cos^{2}(a)-1$. – Kavi Rama Murthy Feb 21 '19 at 8:46

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