Using the steps below, find the value of $x ^ { 2012 } + x ^ { - 2012 }$, where $x + x ^ { - 1 } = \frac { \sqrt { 5 } + 1 } { 2 }$. a) For any real $r$, show that $\left| r + r ^ { - 1 } \right| \geq 2$. What does this tell you about the given $x$? b) Show that $\cos \left( \frac { \pi } { 5 } \right) = \frac { \sqrt { 5 } + 1 } { 4 }$, e.g. compare $\sin \left( \frac { 2 \pi } { 5 } \right)$ and $\sin \left( \frac { 3 \pi } { 5 } \right)$. c) Combine conclusions of parts a and b to express $x$ and therefore the desired quantity in a suitable form.

On the coordinate plane, let $\theta _ { 1 }$ be the acute angle that the line $y = m x ( 0 < m < \sqrt { 3 } )$ makes with the $x$-axis, and let $\theta _ { 2 }$ be the acute angle that the line $y = m x$ makes with the line $y = \sqrt { 3 } x$. What is the value of $m$ that maximizes $3 \sin \theta _ { 1 } + 4 \sin \theta _ { 2 }$? [4 points]
(1) $\frac { \sqrt { 3 } } { 6 }$
(2) $\frac { \sqrt { 3 } } { 7 }$
(3) $\frac { \sqrt { 3 } } { 8 }$
(4) $\frac { \sqrt { 3 } } { 9 }$
(5) $\frac { \sqrt { 3 } } { 10 }$

19. (This question is worth 12 points)

Let $A , B , C$ be the interior angles of $\triangle A B C$. $\tan A , \tan B$ are the two real roots of the equation $x ^ { 2 } + \sqrt { 3 } p x - p + 1 = 0 ( p \in R )$. (1) Find the size of $C$; (2) If $A B = 3 , A C = \sqrt { 6 }$, find the value of $p$.

19. As shown in the figure, $A$, $B$, $C$, $D$ are the four interior angles of quadrilateral $ABCD$.
(1) Prove: $\tan \frac { A } { 2 } = \frac { 1 - \cos A } { \sin A }$;
(2) If $A + C = 180 ^ { \circ }$, $AB = 6$, $BC = 3$, $CD = 4$, $AD = 5$, find the value of $\tan \frac { A } { 2 } + \tan \frac { B } { 2 } + \tan \frac { C } { 2 } + \tan \frac { D } { 2 }$. [Figure]

The vertex of angle $\alpha$ is at the origin, its initial side coincides with the positive $x$-axis, and two points on its terminal side are $A ( 1 , a )$ and $B ( 2 , b )$. If $\cos 2 \alpha = \frac { 2 } { 3 }$, then $| a - b | =$
A. $\frac { 1 } { 5 }$
B. $\frac { \sqrt { 5 } } { 5 }$
C. $\frac { 2 \sqrt { 5 } } { 5 }$
D. (incomplete)

Given $0 < a < \pi$, $\cos \frac{a}{2} = \frac{\sqrt{5}}{5}$, then $\sin\left(a - \frac{\pi}{4}\right) = $
A. $\frac{\sqrt{2}}{10}$
B. $\frac{\sqrt{2}}{5}$
C. $\frac{3\sqrt{2}}{10}$
D. $\frac{7\sqrt{2}}{10}$

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$, and set $\theta _ { k } := ( 2 k + 1 ) \dfrac { \pi } { p }$. We admit that for all $0 \leq k \leq \lfloor p / 2 \rfloor - 1$, $$\int _ { 0 } ^ { 1 } F _ { k } ( t ) d t = \cos \left( q \theta _ { k } \right) \ln \left( 2 \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) - \frac { \pi } { 2 p } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right)$$
Deduce from the previous questions that, for all $( p , q ) \in E _ { 3 }$, $$S _ { p , q } = \frac { 1 } { p } \left( \frac { \pi } { p } \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right) - 2 \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \cos \left( q \theta _ { k } \right) \ln \left( \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) \right)$$

Using the formula established for $( p , q ) \in E _ { 3 }$: $$S _ { p , q } = \frac { 1 } { p } \left( \frac { \pi } { p } \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right) - 2 \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \cos \left( q \theta _ { k } \right) \ln \left( \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) \right)$$ where $\theta_k := (2k+1)\dfrac{\pi}{p}$, deduce the exact values of $S _ { 2,1 }$ and $S _ { 3,1 }$.

Let $\alpha$ and $\beta$ be real numbers such that $- \frac { \pi } { 4 } < \beta < 0 < \alpha < \frac { \pi } { 4 }$. If $\sin ( \alpha + \beta ) = \frac { 1 } { 3 }$ and $\cos ( \alpha - \beta ) = \frac { 2 } { 3 }$, then the greatest integer less than or equal to
$$\left( \frac { \sin \alpha } { \cos \beta } + \frac { \cos \beta } { \sin \alpha } + \frac { \cos \alpha } { \sin \beta } + \frac { \sin \beta } { \cos \alpha } \right) ^ { 2 }$$
is $\_\_\_\_$ .

Let $\frac { \pi } { 2 } < x < \pi$ be such that $\cot x = \frac { - 5 } { \sqrt { 11 } }$. Then
$$\left( \sin \frac { 11 x } { 2 } \right) ( \sin 6 x - \cos 6 x ) + \left( \cos \frac { 11 x } { 2 } \right) ( \sin 6 x + \cos 6 x )$$
is equal to
(A) $\frac { \sqrt { 11 } - 1 } { 2 \sqrt { 3 } }$
(B) $\frac { \sqrt { 11 } + 1 } { 2 \sqrt { 3 } }$
(C) $\frac { \sqrt { 11 } + 1 } { 3 \sqrt { 2 } }$
(D) $\frac { \sqrt { 11 } - 1 } { 3 \sqrt { 2 } }$

If $\tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \tan^3\left(\frac{\pi}{4} + \frac{\alpha}{2}\right)$, then $\sin\theta = $:
(1) $\frac{\sin\alpha(3 + \sin^2\alpha)}{1 + 3\sin^2\alpha}$
(2) $\frac{\sin\alpha(3 - \sin^2\alpha)}{1 + 3\sin^2\alpha}$
(3) $\frac{\sin\alpha(3 + \cos^2\alpha)}{1 + 3\cos^2\alpha}$
(4) $\frac{\sin\alpha(3 - \cos^2\alpha)}{1 + 3\cos^2\alpha}$

Let $\tan ^ { - 1 } y = \tan ^ { - 1 } x + \tan ^ { - 1 } \left( \frac { 2 x } { 1 - x ^ { 2 } } \right)$, where $| x | < \frac { 1 } { \sqrt { 3 } }$. Then a value of $y$ is
(1) $\frac { 3 x + x ^ { 3 } } { 1 + 3 x ^ { 2 } }$
(2) $\frac { 3 x - x ^ { 3 } } { 1 - 3 x ^ { 2 } }$
(3) $\frac { 3 x + x ^ { 3 } } { 1 - 3 x ^ { 2 } }$
(4) $\frac { 3 x - x ^ { 3 } } { 1 + 3 x ^ { 2 } }$

If $0 < x < \frac { 1 } { \sqrt { 2 } }$ and $\frac { \sin ^ { - 1 } x } { \alpha } = \frac { \cos ^ { - 1 } x } { \beta }$, then a value of $\sin \frac { 2 \pi \alpha } { \alpha + \beta }$ is
(1) $4 \sqrt { 1 - x ^ { 2 } } \left( 1 - 2 x ^ { 2 } \right)$
(2) $4 x \sqrt { 1 - x ^ { 2 } } \left( 1 - 2 x ^ { 2 } \right)$
(3) $2 x \sqrt { 1 - x ^ { 2 } } \left( 1 - 4 x ^ { 2 } \right)$
(4) $4 \sqrt { 1 - x ^ { 2 } } \left( 1 - 4 x ^ { 2 } \right)$

$\tan \left( 2 \tan ^ { - 1 } \frac { 1 } { 5 } + \sec ^ { - 1 } \frac { \sqrt { 5 } } { 2 } + 2 \tan ^ { - 1 } \frac { 1 } { 8 } \right)$ is equal to:
(1) 1
(2) 2
(3) $\frac { 1 } { 4 }$
(4) $\frac { 5 } { 4 }$