why is the square of this matrix with sin and cos equal to the identity matrix? I have a question about why the square of the 
matrix  Q, below, is equal to the identity matrix.
Q = 
   cos X  -sin X
   sin X   cos X

My knowledge of trigonometry seems to have rotted away, and I can't 
figure out what rules are used to justify the statement  made
by professor Strang at 11:55
of this video >
Lec 20 | MIT 18.06 Linear Algebra, Spring 2005
(he says the square of  Q is equal to I)
The thing that has me stuck is the following:   In order to calculate the 
first cell of  Q * Q   we take the dot product of the first row and the 
first column.  That would be 
  cos^2 Q  -  sin^2 Q

But how is that equal to the "1" that we want in the upper left most cell 
of the identity matrix ?   
If this were 
      cos^2 Q  +  sin^2 Q
then it would be equal to 1.. but we have a difference here, not a sum.
Thanks in advance !
   chris
epilogue:  thanks for the answers ! I don't know how I missed the fact that he was not making this claim for Q^2,  but  Q-transpose * Q, instead. Appreciate your pointing it out.
 A: The square of this matrix is not equal to the identity matrix!  We compute:
$Q^2 = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} = \begin{bmatrix}  \cos^2 x -  \sin^2 x & -2\sin x \cos x \\ 2 \sin x \cos x & \cos^2 x - \sin^2 x \end{bmatrix}, \tag{1}$
in fact this is
$Q^2 = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}; \tag{2}$
but if we try $Q^TQ$, we get
$Q^TQ = \begin{bmatrix}  \cos^2 x + \sin^2 x & -\cos x \sin x + \cos x \sin x \\ -\sin x \cos x + \sin x \cos x & \cos^2 x+ \sin^2 x \end {bmatrix} = I, \tag{3}$
$I$ being the $2 \times 2$  identity matrix.  
So someone, somewhere must have gotten $Q^2$ and $Q^TQ$ mixed up!
Hope this helps clear things up!  
Cheers!
A: Look at the video you referenced around 21:30, he is not saying that the square of this matrix is the identity matrix. He is saying that if the columns of $Q$ are orthonormal, then $QQ^T=I$ provided the matrix is square. That is to say the matrix is an $n \! \times \! n$ matrix.
Concisely: If $Q$ is a square ($n \! \times \! n$) matrix with orthonormal column vectors, then the product of $Q$ and its transpose equals the identity. 
Also note that the column vectors of your matrix are orthonormal as their dot product is zero.
A: $$ \color{magenta}{
\left(%
\begin{array}{cr}
\cos\left(X\right) & \sin\left(X\right)
\\
\sin\left(X\right) & -\cos\left(X\right)
\end{array}\right)}
$$
really does square to give the identity.
For an $n$ by $n$ matrix with real entries, call it $A,$ if $A$ is symmetric and $A^2 = I$ and $\operatorname{trace} A = n-2,$ then $A$ is a reflection. This applies to the example above with $n=2.$ The eigenvalues are, indeed, $-1,1.$ 
If we have real numbers $\alpha, \beta, \gamma$ such that
$$ \color{blue}{ \alpha^2 + \beta^2 + \gamma^2 = 1    }, $$  then
$$ \color{green}{
\left(
\begin{array}{ccc}
1 - 2 \alpha^2 & - 2 \alpha \beta & -2 \alpha \gamma  \\
- 2 \alpha \beta & 1 - 2 \beta^2 & - 2 \beta \gamma   \\
-2 \alpha \gamma  & - 2 \beta \gamma & 1 - 2 \gamma^2
\end{array}\right)}
$$
is a reflection. It negates the vector with entries $(\alpha, \beta, \gamma)$ and fixes vectors orthogonal to that, such as $(\beta, -\alpha, 0)$ and $(\alpha \gamma, \beta \gamma, \gamma^2 - 1).$ So the eigenvalues are $-1,1,1$ and the trace is $1 = 3-2.$
Oh, if you simply negate all the entries in the 3 by 3  matrix, what you get fixes $(\alpha, \beta, \gamma)$ and rotates in the orthogonal plane by $180^\circ$
Given a column vector $v$ with transpose $v^T$ a row vector, we know that $v^T v$ is the 1 by 1 matrix whose only entry is $v \cdot v = |v|^2.$ In contrast, $v v^T$ is a symmetric $n$ by $n$ matrix of rank $1.$ If, in addition, $|v| = 1,$ then the matrix of the reflection in $v$ is
$$ \color{magenta}{ I - 2 v v^T}.    $$ So, given any $w$ orthogonal to $v,$ we get $v^T w = ( v \cdot w) = (0);$ so $$(I - 2 v v^T) w = Iw - 2 v (v^T w) = w - 2 v (0) = w.$$ But $$(I - 2 v v^T) v = Iv - 2 v (v^T v) = v - 2 v (1) = -v.$$ Meanwhile,
 $$(I - 2 v v^T)^2 = I - 4 v v^T + 4 v v^T v v^T  = I - 4 v v^T + 4 v (v^T v) v^T = I.$$
If $  a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 = 1, $ the reflection that negates $(a,b,c,d,e,f,g,h)$ is 
$$
\color{magenta}{\left(
\begin{array}{cccccccc}
1 - 2 a^2 & -2ab & -2ac &  -2ad & -2ae & -2af & -2ag & -2ah \\
-2ab & 1-2b^2 & -2bc &  -2bd & -2be & -2bf & -2bg & -2bh \\
-2ac & -2bc & 1-2c^2 &  -2cd & -2ce & -2cf & -2cg & -2ch \\
-2ad & -2bd & -2cd &  1-2d^2 & -2de & -2df & -2dg & -2dh \\
-2ae & -2be & -2ce &  -2de & 1-2e^2 & -2ef & -2eg & -2eh \\
-2af & -2bf & -2cf &  -2df & -2ef & 1-2f^2 & -2fg & -2fh \\
-2ag & -2bg & -2cg &  -2dg & -2eg & -2fg & 1-2g^2 & -2gh \\
-2ah & -2bh & -2ch &  -2dh & -2eh & -2fh & -2gh & 1-2h^2 
\end{array}\right)}
$$
with trace 6.
A: Your matrix is the rotation matrix of any vector of length 1 about the origin. Why would that square end up being the identity matrix? It is, however for a selective number of values, as Daniel pointed out.
A: This matrix $Q$ represents a rotation around origin by $X$. Then $Q^2$ represents two consecutive rotation which means a rotation by $2x$:
$$
Q^2=\left[
\begin{array} {ll}
\cos(2x)&-\sin(2x)\\
\sin(2x)&\cos(2x)
\end{array}
\right]
$$
So $Q^2$ is not identity. However if you rotate back by $-x$ then you expect to go back to your original point. So the true inverse of $Q$ is:
$$
Q^*=\left[
\begin{array} {ll}
\cos(x)&\sin(x)\\
-\sin(x)&\cos(x)
\end{array}
\right]
$$
which is indeed its transpose.
A: $$
\left(%
\begin{array}{cr}
\cos\left(X\right) & -\sin\left(X\right)
\\
\sin\left(X\right) & \cos\left(X\right)
\end{array}\right)
=
\cos\left(X\right) - {\rm i}\sin\left(X\right)\,\sigma_{y}\,,
\quad
\sigma_{y}
\equiv
\left(%
\begin{array}{cr}
0 & -{\rm i}
\\
{\rm i} & 0
\end{array}\right)\,,
\quad
\left\vert%
\begin{array}{l}
\mbox{Notice that}
\\
\sigma_{x}^{2} = 2\times 2\ \mbox{identity}.
\end{array}\right.
$$
Then
$$
\left(%
\begin{array}{cr}
\cos\left(X\right) & -\sin\left(X\right)
\\
\sin\left(X\right) & \cos\left(X\right)
\end{array}\right)
=
\cos\left(X\right) - {\rm i}\sin\left(X\right)\,\sigma_{y}
=
{\rm e}^{-{\rm i}X\sigma_{y}}
$$
So,
$$\color{#ff0000}{\large%
\left(%
\begin{array}{cr}
\cos\left(X\right) & -\sin\left(X\right)
\\
\sin\left(X\right) & \cos\left(X\right)
\end{array}\right)^{n}
=
{\rm e}^{-{\rm i}nX\sigma_{y}}
=
\left(%
\begin{array}{cr}
\cos\left(nX\right) & -\sin\left(nX\right)
\\
\sin\left(nX\right) & \cos\left(nX\right)
\end{array}\right)}
$$
