OK, so today I was reading some math books to cure my boredom and then I found a problem containing maximizing values.

Here it is:

Consider real $(x, y).$ Given $x^2 + y^2 = 1$, find $\text{max}((x+y)^2).$

So this is how I solved it: $x^2 + y^2 = (x+y)^2 - 2xy = 1.$

$(x+y)^2 = 2xy + 1$

Now you may ask, how in the world do you solve the right hand side?

Trig comes to the rescue. Let $x = \sin(\theta), y = \cos(\theta)$

$2 \sin(\theta) + \cos(\theta) + 1$

You may recognize this as the double angle identity.

$\sin (2 \theta) + 1$

Obviously, $\text{max}(\sin(\theta)) \Rightarrow \theta = 90^{\circ},$ so it is maximized in our problem at $\theta = 45^{\circ}.$

Thus, the max is $1 + 1 = 2.$

So if you are given a problem considering the maximum of $(x+y)^2$ and you are given $x^2 + y^2,$ trig comes as a very handy shortcut to those problems.

MathProblemSolver101 Jul 29, 2021