If $\frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 4 }$, then $\cos ^ { - 1 } \left( \frac { 12 } { 13 } \cos x + \frac { 5 } { 13 } \sin x \right)$ is equal to
(1) $x - \tan ^ { - 1 } \frac { 4 } { 3 }$
(2) $x + \tan ^ { - 1 } \frac { 4 } { 5 }$
(3) $x - \tan ^ { - 1 } \frac { 5 } { 12 }$
(4) $x + \tan ^ { - 1 } \frac { 5 } { 12 }$

If $\cot x = \frac { 5 } { 12 }$ for some $\mathrm { x } \in \left( \pi , \frac { 3 \pi } { 2 } \right)$ then $\sin 7 x \left( \cos \frac { 13 x } { 2 } + \sin \frac { 13 x } { 2 } \right) + \cos 7 x \left( \cos \frac { 13 x } { 2 } - \sin \frac { 13 x } { 2 } \right)$ is equal to

If $\frac { \cos ^ { 2 } 48 ^ { \circ } - \sin ^ { 2 } 12 ^ { \circ } } { \sin ^ { 2 } 24 ^ { \circ } - \sin ^ { 2 } 6 ^ { \circ } } = \frac { \alpha + \sqrt { 5 } \beta } { 2 }$. Then, the value of $( \alpha + \beta )$ is
(A) 3
(B) 2
(C) 11
(D) 4

Consider the function
$$f ( x ) = \sin 2 x - 3 ( \sin x + \cos x )$$
on the interval $- \dfrac { \pi } { 3 } \leqq x \leqq \dfrac { \pi } { 3 }$.
(1) Let $t = \sin x + \cos x$. Find the range of the values which $t$ can take.
(2) The function $f ( x )$ takes its minimum value $\mathbf { E } - \mathbf { F } \sqrt{\mathbf{G}}$ at $x = \dfrac { \mathbf { H } } { \mathbf { I } }$.

Consider the system of equations
$$\left\{ \begin{array} { l } \cos 2 \alpha + \cos 2 \beta = \frac { 4 } { 15 } \\ \cos \alpha \cos \beta = - \frac { 2 \sqrt { 15 } } { 15 } \end{array} \right.$$
where $0 \leqq \alpha \leqq \pi , \quad 0 \leqq \beta \leqq \pi$, and $\alpha < \beta$ with
$$| \cos \alpha | \geqq | \cos \beta |$$
We want to find the values of $\cos \alpha$ and $\cos \beta$.
Using the double angle formula, from (1) we obtain
$$\cos ^ { 2 } \alpha + \cos ^ { 2 } \beta = \frac { \text { アイ } } { \text { ウエ } }$$
Also, from (2), $\cos ^ { 2 } \alpha \cos ^ { 2 } \beta = \frac { \square } { 15 }$.
Therefore, using condition (3),
$$\cos ^ { 2 } \alpha = \frac { \boldsymbol { \text { カ } }} { \boldsymbol { y } } , \quad \cos ^ { 2 } \beta = \frac { \boldsymbol { \text { ク } }} { \boldsymbol { \text { ケ } } }$$
From (2) and the conditions $0 \leqq \alpha \leqq \pi , 0 \leqq \beta \leqq \pi , \alpha < \beta$,
[2] On the coordinate plane, take point $\mathrm { A } \left( 0 , \frac { 3 } { 2 } \right)$, and on the graph of the function $y = \log _ { 2 } x$, take two points $\mathrm { B } \left( p , \log _ { 2 } p \right), \mathrm { C } \left( q , \log _ { 2 } q \right)$. When point C divides segment AB internally in the ratio $1 : 2$, we want to find the values of $p$ and $q$.
By the condition on the domain of the logarithm, $p >$ タ, $q >$ タ. Here, for a logarithm $\log _ { a } b$, $a$ is called the base and $b$ is called the argument.
The coordinates of the point that divides segment AB internally in the ratio $1 : 2$ are expressed in terms of $p$ as
Since this coincides with the coordinates of C,
holds.
Equation (5) can be rewritten as
$$p = \frac { \square } { \square } q \text { 衣 }$$
Solving the system of equations formed by (4) and (6), and noting that $p >$ □タ, $q >$ □タ, we have
$$p = \square \sqrt { \square } , \quad q = \square \sqrt { \square }$$
The $y$-coordinate of C is $\log _ { 2 }$ (□ヒ $\sqrt { }$ □フ). When this value is rounded to one decimal place (rounding the second decimal place), we get □ヘ. Choose the correct answer for □ヘ from the following options (0) through (b). Here, $\log _ { 10 } 2 = 0.3010$ , $\log _ { 10 } 3 = 0.4771 , \log _ { 10 } 7 = 0.8451$.
(0) 0.3 (1) 0.6 (2) 0.9 (3) 1.3 (4) 1.6 (5) 1.9 (6) 2.3 (7) 2.6 (8) 2.9 (9) 3.3 (a) 3.6 (b) 3.9

Consider the function $$f(x) = \sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3}$$ on the interval $0 \leqq x \leqq \pi$. We are to show that $f(x) > 0$ on $0 < x < \pi$, and to find the area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis.
(1) For $\mathbf{K}$, $\mathbf{N}$, $\mathbf{Q}$, $\mathbf{R}$ in the following sentences, choose the correct answer from the following two choices: (0) increasing, (1) decreasing, and for the other blanks, enter the correct number.
When we differentiate $f(x)$, we have $$f'(x) = (\mathbf{A}\cos^2 x - \mathbf{B})(\mathbf{C}\cos x + \mathbf{D}).$$ Hence, over the range $0 \leqq x \leqq \pi$, there are three $x$'s at which $f'(x) = 0$, and when they are arranged in ascending order, they are given accordingly.
Next, looking at whether $f(x)$ is increasing or decreasing, the behaviour is described accordingly.
Also, we have $$f(0) = 0, \quad f(\pi) = 0, \quad f\left(\frac{\mathbf{L}}{\mathbf{M}}\pi\right) = \frac{\sqrt{\mathbf{S}}}{\mathbf{T}} > 0.$$ Hence we see that $f(x) > 0$ on $0 < x < \pi$.
(2) The area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis is $$S = \frac{\mathbf{UV}}{\mathbf{W}}.$$

$x$ satisfies the simultaneous equations
$$\sin 2x + \sqrt{3}\cos 2x = -1$$
and
$$\sqrt{3}\sin 2x - \cos 2x = \sqrt{3}$$
where $0^{\circ} \leq x \leq 360^{\circ}$. Find the sum of the possible values of $x$.

$\cos \mathrm { x } = \frac { - 4 } { 5 }$ Given this, what is $\cos 2 \mathrm { x }$?
A) $\frac { 3 } { 5 }$
B) $\frac { 5 } { 13 }$
C) $\frac { 12 } { 13 }$
D) $\frac { 24 } { 25 }$
E) $\frac { 7 } { 25 }$

$$\cos x \cdot \cos 2x = \frac { 1 } { 16 \sin x }$$
Given this, what is $\sin 4x$?
A) $\frac { 1 } { 2 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 1 } { 4 }$
D) $\frac { \sqrt { 2 } } { 2 }$
E) $\frac { \sqrt { 3 } } { 2 }$

Given that $\alpha , \beta \in \left[ 0 , \frac { \pi } { 2 } \right]$,
$$\sin ( \alpha - \beta ) = \sin \alpha \cdot \cos \beta$$
Which of the following is true?
A) $\alpha = 0$ or $\beta = \frac { \pi } { 2 }$
B) $\alpha = 0$ or $\beta = \frac { \pi } { 4 }$
C) $\alpha = \frac { \pi } { 2 }$ or $\beta = 0$
D) $\alpha = \frac { \pi } { 2 }$ or $\beta = \frac { \pi } { 2 }$
E) $\alpha = \frac { \pi } { 4 }$ or $\beta = 0$

$\cos x = \frac { \sqrt { 5 } } { 3 }$
Accordingly, I. $\sin \mathrm { x }$ II. $\sin 2 x$ III. $\cos 2 x$ Which of the following values equals a rational number?
A) Only I
B) Only III
C) I and II
D) I and III
E) II and III

For every real number $x$, the number $A$ is defined as $$\sum _ { k = 2 } ^ { 4 } \cos ( 2 k x ) = A$$ Accordingly, $$\sum _ { k = 2 } ^ { 4 } \cos ^ { 2 } ( k x )$$ What is the equivalent of the expression in terms of A?\ A) $A + 2$\ B) $A + 4$\ C) $\frac { \mathrm { A } + 1 } { 2 }$\ D) $\frac { A + 2 } { 2 }$\ E) $\frac { A + 3 } { 2 }$

$$\frac{\cos(2x+y) + \sin(2x-y)}{\cos(2x) + \sin(2x)}$$
Which of the following is the simplified form of this expression?
A) $\cos y - \sin y$ B) $\cos y + \sin y$ C) $\cos x - \sin y$ D) $\sin x - \cos y$ E) $\sin x - \cos x$