In a networking textbook, I've been given the identity:

$$2\cos^2(2\pi f_ct) = 1 + \cos(4\pi f_ct)$$

I can see that this is just a slight shifting of the double angle identity.

$$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta= 2\cos^2(\theta) - 1$$

But in an example problem, the book shows an output given 2 cosines multiplied with slightly different phases.

$$2\cos(2\pi f_ct)\cos(2\pi f_ct + \phi) = $$

$$\cos(\phi) + \cos(4\pi f_ct + \phi)$$

I don't understand how the last step is made. Is there a more general form of the identity?

  • 1
    $\begingroup$ Just to simplify the notation a bit, note that what you're really asking for a proof of is $$2\cos(x)\cos(x+y) = \cos(2x+y)+\cos(y)$$ Hint: Try the product-to-sum rule. $\endgroup$ – user137731 Feb 20 '17 at 1:01

Note that $$\cos (\theta+\phi)=\cos \theta \cos \phi-\sin \theta \sin \phi$$$$\cos (\theta - \phi)=\cos \theta \cos \phi+\sin \theta \sin \phi$$Add both sides to get

$$2\cos \theta \cos \phi =\cos (\theta-\phi)+\cos (\theta+\phi)$$ Known as the product-to-sum rule. Your formula can be derived from here.

  • $\begingroup$ Ah, I'm an idiot. My reference page lists the identity for $cos(\theta)cos(\phi)$, and I didn't link the missing division by 2 with the coefficient of 2 in my case. $\endgroup$ – krb686 Feb 20 '17 at 1:03
  • $\begingroup$ @krb686 Glad to help. $\endgroup$ – S.C.B. Feb 20 '17 at 1:03

Duplication formula: $$\cos2\theta=2\cos^2\theta -1,$$ from which you deduce the linearisation formula: $$\cos^2\theta=\frac12(1-\cos2\theta).$$


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