# Tag Archives: squaring

## An unexpected identity

Typical high school mathematics considers only two types of transformations:  translations & scale changes.  From a broader perspective, a transformation is an operation that changes its input into an output.  A function is also an operation that changes inputs.  Therefore, functions and transformations are the same thing from different perspectives.  This post will explore an unexpected discovery Nurfatimah Merchant and I made when applying the squaring function (transformation) to trigonometric functions, an idea we didn’t fully realize until after the initial publication of PreCalculus Transformed.

When a function is transformed, some points are unchanged (invariant) while others aren’t.  But what makes a point invariant in a transformation?  From a function perspective, point a is invariant under transformation T if $T(a)=a$.  Using this, a squaring transformation is invariant for an input, a, when $a^2=a\Rightarrow a*(a-1)=0 \Rightarrow a=\{0,1\}$.

Therefore, input values of  0 and 1 are invariant under squaring, and all other inputs are changed as follows.

• Negative inputs become positive,
• $a^2 for any $0, and
• $a^2>a$ for any $a>1$.

So what happens when the squaring transformation is applied to the graph of $y=sin(x)$ (the input) to get the graph of $y=(sin(x))^2$ (the output)?  Notice that the output of $sin(x)$ is the input to the squaring transformation, so we are transforming y values.  The invariant points in this case are all points where $y=0$ or $y=1$.  Because squaring transforms all negative inputs into positive outputs, the first image shows a dashed graph of $y=sin(x)$ with the invariant points marked as black points and the negative inputs made positive with the absolute value function. All non-invariant points on $y=|sin(x)|$ have magnitude<1 and become smaller in magnitude when squared, as noted above.  Because the original x-intercepts of $y=sin(x)$ are all locally linear, squaring these creates local “bounce” x-intercepts on the output function looking locally similar to the graphs of polynomial double roots.  The result is shown below. While proof that the final output is precisely another sinusoid comes later, the visual image is very compelling.  This looks like a graph of $y=cos(x)$ under a simple scale change ( $S_{0.5,-0.5}$) and translation ( $T_{0,0.5}$), in that order, giving equation $\displaystyle\frac{y-0.5}{-0.5}=cos(\frac{x}{0.5})$ or $y=\frac{1}{2}-\frac{1}{2}cos(2x)$.  Therefore, $sin^2(x)=\frac{1}{2}-\frac{1}{2}cos(2x)$.

We later rewrote this equation to get $cos(2x)=1-2sin^2(x)$.

The initial equation was a nice enough exercise, but what we realized in the rewriting was that we had just “discovered” the half-angle identity for sine and a double-angle identity for cosine using a graphical variation on the squaring transformation!  No manipulation of angle sum identities was required!  (OK, they really are for an honest  proof, but this is pretty compelling evidence.)

Apply the squaring transformation to the graph of $y=cos(x)$ and you get the half-angle identity for cosine and another variation on the double-angle identity for cosine.

We thought this was a nice way to sneak up on trigonometric identities.  Enjoy.