What significance do squares of trigonometric ratios have algebraically and geometrically? I recently came upon a problem which asked me to:
"Show that $(\cos x + 1)^2 = \frac{\cos2x}{2} + 2\cos x + \frac32$ and hence determine $\int(\cos x + 1)^2 dx$."; 
from Mathematics for the International Student-IB Diploma: SL. Being given a value with which to substitute the argument of the integral meant to be found, the second task proved manageable:
Since $(\cos x + 1)^2 = \frac{\cos2x}{2} + 2\cos x + \frac32$, it is evident that $\int(\cos x + 1)^2 dx = \int(\frac{\cos2x}{2} + 2\cos x + \frac32)dx$, which is solved quite easily. 
However, I struggled at finding a way to show that the initial argument made sense. Here is the work I did:
$$\frac{\cos 2x}{2} + 2 \cos x + \frac32 = (\cos x + 1)^2$$
$$\frac{\cos 2x}{2} + 2 \cos x + \frac32= \cos^2 x + 2 \cos x + 1$$
$$\frac{\cos 2x + 3}{2} = \cos^2 x + 1$$
$$\cos 2x + 3 = 2 \cos^2 x + 2$$
$$\cos 2x = 2\cos^2 x - 1$$
It is at this point that I get stuck. I realize that trigonometric identities will facilitate the above statement, but I do not know which ones "fit". Therefore, I come to my question: what significance do squares of trigonometric ratios have algebraically and geometrically?
 A: I usually tell my students not to memorize a bunch of identities, then very quickly tell them that there are two very useful and important trigonometric identities that let you derive all the rest:


*

*The Pythagorean Identity: $\cos(\theta)^2 + \sin(\theta)^2 = 1$ for all $\theta$, and

*The Angle Addition Formula for Cosine: $\cos(\theta+\varphi) = \cos(\theta)\cos(\varphi) - \sin(\theta)\sin(\varphi)$


Writing $2x = x+x$ and applying the angle addition formula, we get
\begin{equation}
\cos(2x)
= \cos(x+x)
= \cos(x)\cos(x) - \sin(x)\sin(x)
= \cos(x)^2 - \sin(x)^2.\end{equation}
Then the Pythagorean identity implies that $\sin(x)^2 = 1-\cos(x)^2$, and so we have
\begin{equation}
\cos(2x)
= \cos(x)^2 - \sin(x)^2
= \cos(x)^2 - (1-\cos(x)^2)
= 2\cos(x)^2 - 1,\end{equation}
which is what you wanted.

Geometrically, the Pythagorean identity is just the distance formula, which is really just the Pythagorean theorem in disguise.  Suppose that $p = (x_1,y_1)$ and $q = (x_2,y_2)$ are points in the plane.  How do we measure the distance between them?  This is the same as asking how long the segment joining the to points is, and this segment is the hypotenuse of a right triangle with legs of length $|x_1-x_2|$ (the horizontal distance between the two points) and $|y_1-y_2|$ (the vertical distance between the two points).  By the Pythagorean theorem
\begin{equation}
\text{hypotenuse}^2 = \text{leg}_1^2 + \text{leg}_2^2.
\end{equation}
That is
\begin{equation}
d(p,q)^2 = |x_1-x_2|^2 + |y_1-y_2|^2,
\end{equation}
where $d(p,q)$ denotes the distance between the two points.  Now, remember that if $\theta$ is some angle, then $\cos(\theta)$ and $\sin(\theta)$ are the coordinates of a point on the unit circle (specifically, the point where the angle $\theta$ crosses the unit circle).  But if $(x,y)$ is any point on the unit circle, then
\begin{equation}
1 = 1^2
= d((0,0),(x,y))^2
= |0-x|^2 + |0-y|^2
= x^2 + y^2.
\end{equation}
Setting $x=\cos(\theta)$ and $y=\sin(\theta)$, we get the Pythagorean identity.
The angle addition formula also encapsulates some geometric meaning.  There is a nice visualization here.
A: You can derive the difference to obtain $\sin(2x) + 2\sin (x) - 2\sin(x)(\cos(x) + 1) = \sin(2x) - 2\sin(x)\cos(x) = 0$ and then observe that the value at $0$ is the same. 
A: Expanding the expression we have $$(\cos x + 1)^2 = \cos ^2 x + 2 \cos x + 1$$
However, $\cos^2 x = \frac {\cos 2x + 1}{2}$ so we have $$\frac {\cos 2x + 1}{2} + 2 \cos x + 1$$ Factoring $1/2$ from the first term $\frac {\cos 2x + 1}{2}$, we then have $$\frac {\cos 2x}{2} + \frac {1}{2} + 2 \cos x + 1 $$ or $$\frac {\cos 2x}{2} + 2 \cos x + \frac {3}{2}$$
The integral for $\int (\frac {\cos 2x}{2} + 2 \cos x + \frac {3}{2}) dx$ is $$\frac {\sin 2x}{4} + 2 \sin x + \frac {3x}{2} + C$$
