Remembering One Trig Identity « Calculus: Problems and Solutions

Remembering One Trig Identity

July 17, 2009 · Leave a Comment

We’ve been talking about a way to figure out trig identities without having to just memorize them. Here is how I showed the identity

$\displaystyle \cos^2 x = \frac{\cos 2x + 1}{2}$ ,

by just knowing Euler’s formula and some facts on complex numbers. I started off by using Euler’s formula to write

$\displaystyle \cos^2 x = \left(\Re e^{ix}\right)^2$ .

Then I used the fact from complex numbers that $\left(\Re z\right)^2 = \Re(z^2)+\left(\Im z\right)^2$ to say that

$\displaystyle \left(\Re e^{ix}\right)^2 = \Re\left( (e^{ix})^2\right) + \left(\Im e^{ix}\right)^2$
$\displaystyle = \cos 2x + \sin^2 x$ .

Since $\sin^2 x = 1-\cos^2 x$ , this is in turn equal to

$\displaystyle \cos 2x + 1-\cos^2 x$ .

Now we can add $\cos^2 x$ to get

$2\cos^2 x = \cos 2x + 1$ ,

which is pretty much what we wanted. Cool, huh?

It turns out that you can figure out pretty much any of the identities you’re gonna see in calculus or differential equations by just using this technique.

Can you figure out how to show these identities?

$\sin^2 x = \frac{1-\cos 2x}{2}$
$\cos 2x = \cos^2 x - \sin^2 x$
For practice, try to figure this one out without looking up what the right answer is. $\cos (x+y) = ??$

Categories: calculus
Tagged: complex numbers, identities, trig

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Calculus: Problems and Solutions