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.

  • $\begingroup$ 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? $\endgroup$ – Jenny B Feb 21 at 8:43
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    $\begingroup$ 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. $\endgroup$ – Eevee Trainer Feb 21 at 8:50
  • $\begingroup$ ... and the outer two are derived from the one you know by using the Pythagorean identity. $\endgroup$ – NickD Feb 21 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 $a= \frac {\pi} 6$

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

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


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