Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities Questions in English

(C) Let $f(\theta) = 8 \sec^2 \theta + 2 \cos^2 \theta$.
Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms:
$\frac{8 \sec^2 \theta + 2 \cos^2 \theta}{2} \ge \sqrt{8 \sec^2 \theta \times 2 \cos^2 \theta}$
$\frac{8 \sec^2 \theta + 2 \cos^2 \theta}{2} \ge \sqrt{16 \sec^2 \theta \cos^2 \theta}$
Since $\sec^2 \theta \cos^2 \theta = 1$,we have:
$\frac{8 \sec^2 \theta + 2 \cos^2 \theta}{2} \ge \sqrt{16}$
$8 \sec^2 \theta + 2 \cos^2 \theta \ge 2 \times 4$
$8 \sec^2 \theta + 2 \cos^2 \theta \ge 8$.
Thus,the minimum value is $8$.

(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \ge \sqrt{ab}$.
Let $a = \sqrt{\frac{x}{y}}$ and $b = \sqrt{\frac{y}{x}}$.
Then $\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) \ge \sqrt{\sqrt{\frac{x}{y}} \cdot \sqrt{\frac{y}{x}}} = \sqrt{1} = 1$.
Since $\sin \theta = \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right)$ and we know that $\sin \theta \le 1$ for all $\theta \in \mathbb{R}$,the only possible value for the expression is $1$.
Therefore,$\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right) = 1$.
This implies $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 2$.
Squaring both sides,we get $\frac{x}{y} + \frac{y}{x} + 2 = 4$,which simplifies to $\frac{x}{y} + \frac{y}{x} = 2$.
Multiplying by $xy$,we get $x^2 + y^2 = 2xy$,or $(x-y)^2 = 0$.
Thus,$x = y$.

(C) Given $\alpha + \beta = \pi \implies \alpha = \pi - \beta \implies \tan \alpha = -\tan \beta$.
Also,$\beta + \gamma = \alpha$.
Taking tangent on both sides: $\tan(\beta + \gamma) = \tan \alpha$.
$\frac{\tan \beta + \tan \gamma}{1 - \tan \beta \tan \gamma} = \tan \alpha$.
Substitute $\tan \beta = -\tan \alpha$:
$\frac{-\tan \alpha + \tan \gamma}{1 - (-\tan \alpha) \tan \gamma} = \tan \alpha$.
$-\tan \alpha + \tan \gamma = \tan \alpha(1 + \tan \alpha \tan \gamma) = \tan \alpha + \tan^2 \alpha \tan \gamma$.
$\tan^2 \alpha \tan \gamma = -2 \tan \alpha + \tan \gamma$.
Since $\tan \alpha = -\tan \beta$,we have $\tan^2 \alpha = \tan \alpha \cdot (-\tan \beta) = -\tan \alpha \tan \beta$.
Substituting $\tan \alpha = -\tan \beta$ into the equation $\tan^2 \alpha \tan \gamma = -2 \tan \alpha + \tan \gamma$:
$\tan^2 \alpha \tan \gamma = 2 \tan \beta + \tan \gamma$.
$\tan^2 \alpha = \frac{2 \tan \beta + \tan \gamma}{\tan \gamma}$.
Since $\alpha = \pi - \beta$ and $\beta, \gamma$ are positive,$\alpha$ is in the second quadrant,so $\tan \alpha$ must be negative.
Therefore,$\tan \alpha = -\sqrt{\frac{2 \tan \beta + \tan \gamma}{\tan \gamma}}$.

(B) Given $E = \frac{25\sec^4 x - 50\sec^2 x + 74}{\tan^2 x}$.
Using the identity $\sec^2 x = 1 + \tan^2 x$,we can rewrite the numerator:
$E = \frac{25(\sec^2 x - 1)^2 + 49}{\tan^2 x}$
Since $\sec^2 x - 1 = \tan^2 x$,we have:
$E = \frac{25(\tan^2 x)^2 + 49}{\tan^2 x}$
$E = 25\tan^2 x + \frac{49}{\tan^2 x}$
Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:
$E \ge 2 \sqrt{25\tan^2 x \cdot \frac{49}{\tan^2 x}}$
$E \ge 2 \sqrt{25 \cdot 49}$
$E \ge 2 \cdot 5 \cdot 7 = 70$
Thus,the minimum value is $70$.

(B) We know that for any real numbers $x$ and $y$,the arithmetic mean is greater than or equal to the geometric mean,i.e.,$\frac{x+y}{2} \geq \sqrt{xy}$.
Squaring both sides,we get $\frac{(x+y)^2}{4} \geq xy$,which implies $\frac{4xy}{(x+y)^2} \leq 1$.
Since $cosec^2 \theta \geq 1$ for all real $\theta$,the equation $cosec^2 \theta = \frac{4xy}{(x+y)^2}$ can only hold if both sides are equal to $1$.
This occurs when $\frac{4xy}{(x+y)^2} = 1$,which simplifies to $4xy = (x+y)^2$,or $(x-y)^2 = 0$,meaning $x = y$.
Additionally,for $cosec^2 \theta$ to be defined,$x$ and $y$ must be non-zero (since $x=y$,$x \neq 0$ implies $y \neq 0$ and $x+y \neq 0$).

(B) Let $E = \sin \theta + \cos \theta + \sin 2\theta$.
Let $t = \sin \theta + \cos \theta$. Then $t^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + \sin 2\theta$,so $\sin 2\theta = t^2 - 1$.
The range of $t = \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$ is $[-\sqrt{2}, \sqrt{2}]$.
Substituting $t$ into the expression,we get $E = t + t^2 - 1 = t^2 + t - 1$.
This is a quadratic in $t$ opening upwards. The maximum value on the interval $[-\sqrt{2}, \sqrt{2}]$ occurs at the boundary $t = \sqrt{2}$.
$E_{\max} = (\sqrt{2})^2 + \sqrt{2} - 1 = 2 + \sqrt{2} - 1 = 1 + \sqrt{2}$.
Note that $1 + \sqrt{2} = \tan(\frac{3\pi}{8})$.

(C) The expression $99 \cos 2\theta - 20 \sin 2\theta$ is of the form $a \cos x + b \sin x$,which lies in the range $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 99$ and $b = -20$.
$\sqrt{99^2 + (-20)^2} = \sqrt{9801 + 400} = \sqrt{10201} = 101$.
Thus,$-101 \leq 99 \cos 2\theta - 20 \sin 2\theta \leq 101$.
For the equation to have a solution,we must have $-101 \leq 20p + 35 \leq 101$.
Subtracting $35$ from all sides: $-136 \leq 20p \leq 66$.
Dividing by $20$: $-6.8 \leq p \leq 3.3$.
The integral values of $p$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3\}$.
The total number of such integral values is $10$.

(B) The given equation is $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.
Rearranging the terms,we get $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.
Multiplying by $-\frac{1}{2}$,we obtain $-\cos(ax + b) = \frac{x^2 - 2x + 5}{4}$.
This can be written as $-\cos(ax + b) = \frac{(x - 1)^2 + 4}{4} = \frac{(x - 1)^2}{4} + 1$.
We know that $-\cos(ax + b) \in [-1, 1]$.
Also,$\frac{(x - 1)^2}{4} + 1 \geq 1$ for all real $x$.
For the equation to have a solution,both sides must be equal to $1$.
Thus,$\frac{(x - 1)^2}{4} + 1 = 1 \Rightarrow x = 1$.
Substituting $x = 1$ into $-\cos(ax + b) = 1$,we get $\cos(ax + b) = -1$.
This implies $ax + b = (2k + 1)\pi$ for some integer $k$.
For $x = 1$,we have $a + b = (2k + 1)\pi$.
Taking $k = 0$,we get $a + b = \pi$.

(B) Given $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$.
Using $1 + \tan^2 x = \sec^2 x$,we have $y = (7 + 6 \tan x - \tan^2 x) \cos^2 x$.
$y = 7 \cos^2 x + 6 \sin x \cos x - \sin^2 x$.
Using $\cos^2 x = \frac{1 + \cos 2x}{2}$,$\sin^2 x = \frac{1 - \cos 2x}{2}$,and $\sin x \cos x = \frac{\sin 2x}{2}$:
$y = 7 \left( \frac{1 + \cos 2x}{2} \right) + 3 \sin 2x - \left( \frac{1 - \cos 2x}{2} \right)$.
$y = \frac{7 + 7 \cos 2x - 1 + \cos 2x}{2} + 3 \sin 2x = 4 \cos 2x + 3 \sin 2x + 3$.
The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^2 + b^2}$.
Here,the maximum value is $\sqrt{4^2 + 3^2} + 3 = 5 + 3 = 8$.
Thus,$\lambda = 8$.
We need to find $\log_{\sqrt{2}}(8) = \log_{\sqrt{2}}((\sqrt{2})^6) = 6$.

(D) Let $p = \sin x + \cos x$. Then $p^2 = 1 + 2 \sin x \cos x$,which implies $\sin x \cos x = \frac{p^2 - 1}{2}$.
We know that $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{p^2 - 1}$.
Also,$\sec x + \csc x = \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x} = \frac{p}{\frac{p^2 - 1}{2}} = \frac{2p}{p^2 - 1}$.
Substituting these into the function: $f(x) = |p + \frac{2}{p^2 - 1} + \frac{2p}{p^2 - 1}| = |p + \frac{2(1 + p)}{p^2 - 1}| = |p + \frac{2(1 + p)}{(p - 1)(p + 1)}| = |p + \frac{2}{p - 1}|$.
We can rewrite this as $f(x) = |(p - 1) + \frac{2}{p - 1} + 1|$.
For $p - 1 > 0$,by $AM$-$GM$ inequality,$(p - 1) + \frac{2}{p - 1} \geq 2\sqrt{(p - 1) \cdot \frac{2}{p - 1}} = 2\sqrt{2}$.
Thus,$f(x) \geq 2\sqrt{2} + 1$.

(B) The range of the function $f(x) = \cos(\tan x)$ is $[-1, 1]$.
Since the sine function $g(u) = \sin u$ is a strictly increasing function on the interval $[-1, 1]$,the maximum value of the composite function $\sin(\cos(\tan x))$ will occur at the maximum value of the inner function $\cos(\tan x)$.
The maximum value of $\cos(\tan x)$ is $1$ (which occurs when $\tan x = 0$).
Therefore,the maximum value of $\sin(\cos(\tan x))$ is $\sin(1)$.

(D) Let $f(x) = \sin x$ for $x \in (0, \pi)$.
Since $f''(x) = -\sin x < 0$ for all $x \in (0, \pi)$,the function $f(x)$ is concave down.
By Jensen's Inequality,for a concave function,$\frac{f(x) + f(y) + f(z)}{3} \leq f\left(\frac{x+y+z}{3}\right)$.
Let $x = A$,$y = B$,and $z = C - \frac{\pi}{2}$.
Then $\frac{\sin A + \sin B + \sin(C - \frac{\pi}{2})}{3} \leq \sin\left(\frac{A + B + C - \frac{\pi}{2}}{3}\right)$.
Since $A + B + C = \pi$,we have $\frac{\sin A + \sin B - \cos C}{3} \leq \sin\left(\frac{\pi - \frac{\pi}{2}}{3}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
Therefore,$\sin A + \sin B - \cos C \leq \frac{3}{2}$.
The maximum value is $\frac{3}{2}$.

(A) Let $S = \cos 2\theta + \cos \theta$.
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$S = 2\cos^2 \theta - 1 + \cos \theta$.
Completing the square for the expression in $\cos \theta$:
$S = 2(\cos^2 \theta + \frac{1}{2}\cos \theta) - 1$.
$S = 2(\cos^2 \theta + \frac{1}{2}\cos \theta + \frac{1}{16} - \frac{1}{16}) - 1$.
$S = 2(\cos \theta + \frac{1}{4})^2 - \frac{1}{8} - 1$.
$S = 2(\cos \theta + \frac{1}{4})^2 - \frac{9}{8}$.
Since the minimum value of $(\cos \theta + \frac{1}{4})^2$ is $0$ (which occurs when $\cos \theta = -1/4$,a valid value for $\cos \theta$),the minimum value of $S$ is $-9/8$.

(A) Let $f(\theta) = 2\sin^2\theta - 3\sin\theta$. Let $x = \sin\theta$,where $x \in [-1, 1]$.
Then the expression becomes $g(x) = 2x^2 - 3x$.
This is a downward-opening parabola if the coefficient of $x^2$ were negative,but here it is upward-opening with vertex at $x = -b/(2a) = -(-3)/(2 \times 2) = 3/4$.
Since $3/4 \in [-1, 1]$,the minimum value occurs at $x = 3/4$:
$g(3/4) = 2(3/4)^2 - 3(3/4) = 2(9/16) - 9/4 = 9/8 - 18/8 = -9/8$.
The maximum value occurs at the boundaries of the interval $x \in [-1, 1]$.
At $x = -1$: $g(-1) = 2(-1)^2 - 3(-1) = 2 + 3 = 5$.
At $x = 1$: $g(1) = 2(1)^2 - 3(1) = 2 - 3 = -1$.
Comparing $5$ and $-1$,the maximum value is $5$.
Thus,the maximum and minimum values are $5$ and $-9/8$ respectively.

(A) We want to find the minimum value of $f(\theta) = 3 \sin \theta + \text{cosec}^3 \theta$ for $0 < \theta < \pi$.
Using the $A.M. \ge G.M.$ inequality for four positive terms: $\sin \theta, \sin \theta, \sin \theta, \text{cosec}^3 \theta$.
$\frac{\sin \theta + \sin \theta + \sin \theta + \text{cosec}^3 \theta}{4} \ge \sqrt[4]{(\sin \theta \cdot \sin \theta \cdot \sin \theta \cdot \text{cosec}^3 \theta)}$
$\frac{3 \sin \theta + \text{cosec}^3 \theta}{4} \ge \sqrt[4]{(\sin^3 \theta \cdot \text{cosec}^3 \theta)}$
Since $\sin \theta \cdot \text{cosec} \theta = 1$,we have:
$\frac{3 \sin \theta + \text{cosec}^3 \theta}{4} \ge \sqrt[4]{1^3} = 1$
$3 \sin \theta + \text{cosec}^3 \theta \ge 4$
Thus,the minimum value is $4$.

(A) Given the equation $\sin^2 \theta = \frac{x^2 + y^2}{2xy}$.
Since $\sin^2 \theta \leq 1$,we must have $\frac{x^2 + y^2}{2xy} \leq 1$.
Also,for $\sin^2 \theta$ to be non-negative,$x$ and $y$ must have the same sign,so $\frac{x^2 + y^2}{2xy} > 0$.
By the Arithmetic Mean-Geometric Mean inequality,for positive $x, y$,$\frac{x^2 + y^2}{2} \geq \sqrt{x^2 y^2} = |xy|$.
Thus,$\frac{x^2 + y^2}{2xy} \geq 1$ when $x, y$ have the same sign.
Since we require $\frac{x^2 + y^2}{2xy} \leq 1$ and $\frac{x^2 + y^2}{2xy} \geq 1$,it must be that $\frac{x^2 + y^2}{2xy} = 1$.
This implies $x^2 + y^2 = 2xy$,which simplifies to $(x - y)^2 = 0$,hence $x = y$.

(A) We know that for $n \geq 1$ and $x \in \mathbb{R}$,$0 \leq \sin^2 x \leq 1$ and $0 \leq \cos^2 x \leq 1$.
Since $0 \leq \sin^2 x \leq 1$,it follows that $0 \leq \sin^{2n} x \leq \sin^2 x$.
Similarly,$0 \leq \cos^{2n} x \leq \cos^2 x$.
Adding these inequalities,we get $0 \leq \sin^{2n} x + \cos^{2n} x \leq \sin^2 x + \cos^2 x$.
Since $\sin^2 x + \cos^2 x = 1$,we have $0 < \sin^{2n} x + \cos^{2n} x \leq 1$.
Specifically,the minimum value is $1/2^{n-1}$ (at $x = \pi/4$) and the maximum value is $1$ (at $x = 0$ or $x = \pi/2$).

(D) Given that $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$.
Since the maximum value of the sine function is $1$,this equation holds true only if $\sin \theta_1 = 1$,$\sin \theta_2 = 1$,and $\sin \theta_3 = 1$.
This implies $\theta_1 = \theta_2 = \theta_3 = \frac{\pi}{2}$.
Now,calculating the sum of cosines:
$\cos \theta_1 + \cos \theta_2 + \cos \theta_3 = \cos(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 0 + 0 + 0 = 0$.

(B) Let $f(\theta) = 12 \sin \theta - 9 \sin^2 \theta$.
To find the maximum value,let $x = \sin \theta$,where $x \in [-1, 1]$.
Then $f(x) = 12x - 9x^2$.
To find the critical points,take the derivative with respect to $x$:
$f'(x) = 12 - 18x$.
Setting $f'(x) = 0$ gives $12 - 18x = 0$,so $x = \frac{12}{18} = \frac{2}{3}$.
Since $\frac{2}{3} \in [-1, 1]$,we evaluate the function at this point:
$f\left(\frac{2}{3}\right) = 12\left(\frac{2}{3}\right) - 9\left(\frac{2}{3}\right)^2 = 8 - 9\left(\frac{4}{9}\right) = 8 - 4 = 4$.
Since $f''(x) = -18 < 0$,the function has a local maximum at $x = \frac{2}{3}$.
Thus,the maximum value is $4$.

(C) We know that $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.
Let $a = \sin^2x$ and $b = \cos^2x$.
Then $K = (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos^2x)$.
Since $\sin^2x + \cos^2x = 1$,we have $K = 1 - 3\sin^2x\cos^2x$.
Multiplying and dividing by $4$,we get $K = 1 - \frac{3}{4}(4\sin^2x\cos^2x) = 1 - \frac{3}{4}(\sin 2x)^2$.
Since $0 \leq \sin^2 2x \leq 1$,we have $0 \leq \frac{3}{4}\sin^2 2x \leq \frac{3}{4}$.
Multiplying by $-1$,we get $-\frac{3}{4} \leq -\frac{3}{4}\sin^2 2x \leq 0$.
Adding $1$ to all parts,we get $1 - \frac{3}{4} \leq 1 - \frac{3}{4}\sin^2 2x \leq 1 + 0$.
Thus,$\frac{1}{4} \leq K \leq 1$.
Therefore,$K \in [\frac{1}{4}, 1]$.

(B) Given expression: $f(\theta) = 5 \cos \theta + 3 \cos \left( \theta + \frac{\pi}{3} \right) - 1$
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$f(\theta) = 5 \cos \theta + 3 \left( \cos \theta \cdot \cos \frac{\pi}{3} - \sin \theta \cdot \sin \frac{\pi}{3} \right) - 1$
Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:
$f(\theta) = 5 \cos \theta + 3 \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) - 1$
$f(\theta) = \left( 5 + \frac{3}{2} \right) \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta - 1$
$f(\theta) = \frac{13}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta - 1$
The expression is of the form $a \cos \theta + b \sin \theta + c$,which has a maximum value of $\sqrt{a^2 + b^2} + c$:
$\text{Maximum value} = \sqrt{\left( \frac{13}{2} \right)^2 + \left( -\frac{3\sqrt{3}}{2} \right)^2} - 1$
$\text{Maximum value} = \sqrt{\frac{169}{4} + \frac{27}{4}} - 1$
$\text{Maximum value} = \sqrt{\frac{196}{4}} - 1$
$\text{Maximum value} = \sqrt{49} - 1 = 7 - 1 = 6$

(C) Given that $A + B + C = \frac{\pi}{2}$.
This implies $A + B = \frac{\pi}{2} - C$.
Taking the tangent on both sides,we get $\tan(A + B) = \tan(\frac{\pi}{2} - C)$.
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(\frac{\pi}{2} - C) = \cot C = \frac{1}{\tan C}$,we have:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1}{\tan C}$.
Cross-multiplying gives:
$\tan C(\tan A + \tan B) = 1 - \tan A \tan B$.
Expanding the left side:
$\tan C \tan A + \tan B \tan C = 1 - \tan A \tan B$.
Rearranging the terms,we get:
$\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.

(A) Let $f(\theta) = 2 \sin^2 \theta + 8 \csc^2 \theta$.
Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for positive terms:
$\frac{2 \sin^2 \theta + 8 \csc^2 \theta}{2} \ge \sqrt{2 \sin^2 \theta \cdot 8 \csc^2 \theta}$
$\frac{2 \sin^2 \theta + 8 \csc^2 \theta}{2} \ge \sqrt{16 \sin^2 \theta \cdot \frac{1}{\sin^2 \theta}}$
$\frac{2 \sin^2 \theta + 8 \csc^2 \theta}{2} \ge \sqrt{16}$
$2 \sin^2 \theta + 8 \csc^2 \theta \ge 2 \times 4$
$2 \sin^2 \theta + 8 \csc^2 \theta \ge 8$.
The equality holds when $2 \sin^2 \theta = 8 \csc^2 \theta$,which implies $\sin^4 \theta = 4$,which is impossible as $\sin^2 \theta \le 1$.
Since $\sin^2 \theta \in (0, 1]$,let $x = \sin^2 \theta$. Then $f(x) = 2x + \frac{8}{x}$.
For $x \in (0, 1]$,the function $f(x) = 2x + \frac{8}{x}$ is a decreasing function because $f'(x) = 2 - \frac{8}{x^2} < 0$ for $x < 2$.
Thus,the minimum value occurs at the largest possible value of $x$,which is $x = 1$.
$f(1) = 2(1) + \frac{8}{1} = 10$.

(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.
Using $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we get:
$f(x) = 4 + \frac{1}{2} [4(1 - \cos^2 x) \cos^2 x] - 2 \cos^4 x$
$f(x) = 4 + 2 \cos^2 x - 2 \cos^4 x - 2 \cos^4 x$
$f(x) = 4 + 2 \cos^2 x - 4 \cos^4 x$.
Let $t = \cos^2 x$,where $t \in [0, 1]$.
Then $g(t) = -4t^2 + 2t + 4$.
This is a downward parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-4)} = \frac{1}{4}$.
Since $\frac{1}{4} \in [0, 1]$,the maximum value $M$ occurs at $t = \frac{1}{4}$:
$M = g(\frac{1}{4}) = -4(\frac{1}{16}) + 2(\frac{1}{4}) + 4 = -\frac{1}{4} + \frac{1}{2} + 4 = \frac{1}{4} + 4 = \frac{17}{4}$.
The minimum value $m$ occurs at the boundaries $t = 0$ or $t = 1$:
$g(0) = 4$
$g(1) = -4(1)^2 + 2(1) + 4 = 2$.
Thus,$m = 2$.
$M - m = \frac{17}{4} - 2 = \frac{17 - 8}{4} = \frac{9}{4}$.

(A) Given $A + B = \frac{\pi}{6}$. Let $y = \tan A + \tan B$.
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\frac{1}{\sqrt{3}} = \frac{y}{1 - \tan A \tan B}$.
Thus,$\tan A \tan B = 1 - \sqrt{3}y$.
Since $A, B > 0$ and $A+B = \frac{\pi}{6}$,both $\tan A$ and $\tan B$ are positive,so $\tan A \tan B > 0$,which implies $1 - \sqrt{3}y > 0$,or $y < \frac{1}{\sqrt{3}}$.
By $AM \ge GM$,we have $\frac{\tan A + \tan B}{2} \ge \sqrt{\tan A \tan B}$.
Substituting $y$,we get $\frac{y}{2} \ge \sqrt{1 - \sqrt{3}y}$.
Squaring both sides,$\frac{y^2}{4} \ge 1 - \sqrt{3}y$,so $y^2 + 4\sqrt{3}y - 4 \ge 0$.
The roots of $y^2 + 4\sqrt{3}y - 4 = 0$ are $y = \frac{-4\sqrt{3} \pm \sqrt{48 + 16}}{2} = -2\sqrt{3} \pm 4$.
Since $y > 0$,we must have $y \ge 4 - 2\sqrt{3}$.

(C) We are given the expression $E = \frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}$.
We can rewrite this expression as $E = \frac{x^m}{1 + x^{2m}} \times \frac{y^n}{1 + y^{2n}}$.
Dividing the numerator and denominator of each fraction by $x^m$ and $y^n$ respectively,we get:
$E = \frac{1}{\left( \frac{1}{x^m} + x^m \right)} \times \frac{1}{\left( \frac{1}{y^n} + y^n \right)}$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for any positive real number $a$,$a + \frac{1}{a} \ge 2$,with equality holding when $a = 1$.
Therefore,$x^m + \frac{1}{x^m} \ge 2$ and $y^n + \frac{1}{y^n} \ge 2$.
This implies that $\frac{1}{x^m + \frac{1}{x^m}} \le \frac{1}{2}$ and $\frac{1}{y^n + \frac{1}{y^n}} \le \frac{1}{2}$.
Multiplying these inequalities,we get the maximum value of $E$ as $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(A) Let $f(\theta) = 3 \cos \theta + 5 \sin \left( \theta - \frac{\pi}{6} \right)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$f(\theta) = 3 \cos \theta + 5 \left( \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6} \right)$
$f(\theta) = 3 \cos \theta + 5 \left( \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta \right)$
$f(\theta) = \frac{5\sqrt{3}}{2} \sin \theta + \left( 3 - \frac{5}{2} \right) \cos \theta$
$f(\theta) = \frac{5\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta$.
The maximum value of $a \sin \theta + b \cos \theta$ is $\sqrt{a^2 + b^2}$.
Maximum value $= \sqrt{\left( \frac{5\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2}$
$= \sqrt{\frac{25 \times 3}{4} + \frac{1}{4}} = \sqrt{\frac{75 + 1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19}$.

(B) The expression is of the form $y = a \sin x + b \cos x$.
The maximum value of this expression is given by $\sqrt{a^2 + b^2}$.
Here,$a = 3$ and $b = 4$.
Therefore,the maximum value is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(A) Given the function $h(x) = \sin(2x) + 5$.
We know that the range of the sine function is $[-1, 1]$,so $-1 \leq \sin(2x) \leq 1$.
Adding $5$ to all parts of the inequality:
$-1 + 5 \leq \sin(2x) + 5 \leq 1 + 5$.
$4 \leq \sin(2x) + 5 \leq 6$.
Thus,the maximum value of $h(x)$ is $6$ and the minimum value of $h(x)$ is $4$.

(A) Given function is $f(x) = |\sin 4x + 3|$.
We know that the range of the sine function is $-1 \leq \sin \theta \leq 1$ for any $\theta$.
Therefore,$-1 \leq \sin 4x \leq 1$.
Adding $3$ to all parts of the inequality:
$-1 + 3 \leq \sin 4x + 3 \leq 1 + 3$
$2 \leq \sin 4x + 3 \leq 4$.
Since the values in the interval $[2, 4]$ are all positive,the absolute value function does not change the range:
$|2| \leq |\sin 4x + 3| \leq |4|$
$2 \leq f(x) \leq 4$.
Thus,the maximum value is $4$ and the minimum value is $2$.

(B) Let $f(x) = \sin x + \cos x$.
To find the maximum value,we can rewrite the function as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right)$.
Since the maximum value of $\sin \theta$ is $1$,the maximum value of $f(x)$ is $\sqrt{2} \times 1 = \sqrt{2}$.

We have $\text{L.H.S.} = \cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$\text{L.H.S.} = \frac{1 + \cos 2x}{2} + \frac{1 + \cos(2x + \frac{2\pi}{3})}{2} + \frac{1 + \cos(2x - \frac{2\pi}{3})}{2}$
$= \frac{1}{2} [3 + \cos 2x + \cos(2x + \frac{2\pi}{3}) + \cos(2x - \frac{2\pi}{3})]$
Using the formula $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x \cos \frac{2\pi}{3}]$
Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x(-\frac{1}{2})]$
$= \frac{1}{2} [3 + \cos 2x - \cos 2x]$
$= \frac{3}{2} = \text{R.H.S.}$

(A) Using the $AM \geq GM$ inequality for two positive numbers $2^{\sin x}$ and $2^{\cos x}$:
$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$
$\Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2^{1 + \frac{\sin x + \cos x}{2}}$
We know that the minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.
Substituting this value:
$\min(2^{\sin x} + 2^{\cos x}) = 2^{1 + \frac{-\sqrt{2}}{2}} = 2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - \frac{1}{\sqrt{2}}}$

(A) The given equation is $3 \sin x + 4 \cos x = k + 1$.
We know that the range of the function $f(x) = a \sin x + b \cos x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.
For the equation to have a solution,the value $k + 1$ must lie within this range:
$-5 \le k + 1 \le 5$
Subtracting $1$ from all sides,we get:
$-6 \le k \le 4$
The integral values of $k$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$.
The total number of such integral values is $11$.

(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.
We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.
So,$f(x) = a^{a^{x}} + \frac{a}{a^{a^{x}}}$.
Since $a > 0$,both $a^{a^{x}}$ and $\frac{a}{a^{a^{x}}}$ are positive real numbers.
By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for any two positive numbers $u$ and $v$,$u + v \geq 2\sqrt{uv}$.
Let $u = a^{a^{x}}$ and $v = \frac{a}{a^{a^{x}}}$.
Then $f(x) = u + v \geq 2\sqrt{u \cdot v} = 2\sqrt{a^{a^{x}} \cdot \frac{a}{a^{a^{x}}}} = 2\sqrt{a}$.
Thus,the minimum value is $2\sqrt{a}$.

(B) Given $\alpha = \max \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$.
We can rewrite the expression as $2^{6 \sin 3x} \cdot 2^{8 \cos 3x} = 2^{6 \sin 3x + 8 \cos 3x}$.
The range of $6 \sin 3x + 8 \cos 3x$ is $[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}] = [-10, 10]$.
Thus,$\alpha = 2^{10}$ and $\beta = 2^{-10}$.
Then $\alpha^{1/5} = (2^{10})^{1/5} = 2^2 = 4$ and $\beta^{1/5} = (2^{-10})^{1/5} = 2^{-2} = 1/4$.
The roots of $8x^2 + bx + c = 0$ are $4$ and $1/4$.
Sum of roots: $4 + 1/4 = 17/4 = -b/8 \Rightarrow b = -34$.
Product of roots: $4 \cdot 1/4 = 1 = c/8 \Rightarrow c = 8$.
Therefore,$c - b = 8 - (-34) = 42$.

(C) Given $f(x) = \cos 5x + A \cos 4x + B \cos 3x + C \cos 2x + D \cos x + E$.
Since $f(x) = f(2\pi - x)$,we have $f\left(\frac{\pi}{5}\right) = f\left(\frac{9\pi}{5}\right)$,$f\left(\frac{2\pi}{5}\right) = f\left(\frac{8\pi}{5}\right)$,$f\left(\frac{3\pi}{5}\right) = f\left(\frac{7\pi}{5}\right)$,and $f\left(\frac{4\pi}{5}\right) = f\left(\frac{6\pi}{5}\right)$.
The expression $T$ can be written as $T = f(0) - 2f\left(\frac{\pi}{5}\right) + 2f\left(\frac{2\pi}{5}\right) - 2f\left(\frac{3\pi}{5}\right) + 2f\left(\frac{4\pi}{5}\right) - f(\pi)$.
Substituting the values of $f(x)$ at these points,the terms involving $A, C, E$ cancel out due to the symmetry of the cosine function around $\pi$.
Specifically,$f(0) - f(\pi) = 2(1 + B + D)$.
The remaining terms also involve only $B$ and $D$ coefficients.
Thus,$T$ depends on $B$ and $D$ but is independent of $A, C, E$.

(A) Let $P = \sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$.
Without loss of generality,assume $0 \leq x \leq y \leq z \leq 1$.
Then $P = \sqrt{y-x} + \sqrt{z-y} + \sqrt{z-x}$.
To maximize $P$,we set $x=0$ and $z=1$,which gives $P = \sqrt{y} + \sqrt{1-y} + 1$.
Let $f(y) = \sqrt{y} + \sqrt{1-y} + 1$ for $y \in [0, 1]$.
Using the substitution $y = \sin^2 \theta$,where $\theta \in [0, \pi/2]$,we get $f(\theta) = \sin \theta + \cos \theta + 1$.
We know that $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$,which has a maximum value of $\sqrt{2}$ at $\theta = \pi/4$.
Thus,the maximum value of $P$ is $\sqrt{2} + 1$.

(D) Given equation: $\lambda = \cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x$
Convert all terms to $\cos x$:
$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x$
Expand the terms:
$\lambda = (4 \cos ^4 x - 4 \cos ^2 x + 1) - 2(1 - 2 \cos ^2 x + \cos ^4 x) - 2 \cos ^2 x$
Simplify:
$\lambda = 4 \cos ^4 x - 4 \cos ^2 x + 1 - 2 + 4 \cos ^2 x - 2 \cos ^4 x - 2 \cos ^2 x$
$\lambda = 2 \cos ^4 x - 2 \cos ^2 x - 1$
Let $t = \cos ^2 x$,where $t \in [0, 1]$:
$f(t) = 2t^2 - 2t - 1$
Find the range of $f(t)$ for $t \in [0, 1]$:
$f'(t) = 4t - 2$. Setting $f'(t) = 0$ gives $t = \frac{1}{2}$.
$f(0) = -1$
$f(1) = 2 - 2 - 1 = -1$
$f(\frac{1}{2}) = 2(\frac{1}{4}) - 2(\frac{1}{2}) - 1 = \frac{1}{2} - 1 - 1 = -\frac{3}{2}$
Thus,the range of $\lambda$ is $[-\frac{3}{2}, -1]$.

(B) Let $x = \cos^2 \theta$,where $x \in [0, 1]$.
Then $\sin^4 \theta = (1 - x)^2 = 1 - 2x + x^2$.
$f(\theta) = \frac{(1 - 2x + x^2) + 3x}{(1 - 2x + x^2) + x} = \frac{x^2 + x + 1}{x^2 - x + 1}$.
Let $y = \frac{x^2 + x + 1}{x^2 - x + 1}$.
$y(x^2 - x + 1) = x^2 + x + 1 \implies x^2(y - 1) - x(y + 1) + (y - 1) = 0$.
For $x$ to be real,the discriminant $D \ge 0$.
$D = (y + 1)^2 - 4(y - 1)^2 \ge 0$.
$(y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \ge 0$.
$(3 - y)(3y - 1) \ge 0 \implies (y - 3)(3y - 1) \le 0$.
So,$y \in [1/3, 3]$.
Thus,$\alpha = 1/3$ and $\beta = 3$.
The common ratio $r = \frac{\alpha}{\beta} = \frac{1/3}{3} = 1/9$.
The sum of the infinite $G.P.$ is $S = \frac{a}{1 - r} = \frac{64}{1 - 1/9} = \frac{64}{8/9} = 64 \times \frac{9}{8} = 72$.

(A) Let $f(\theta) = \sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2 \theta$.
To find the maximum value of $\frac{1}{f(\theta)}$,we need to find the minimum value of $f(\theta)$.
Using trigonometric identities:
$f(\theta) = \frac{1 - \cos 2\theta}{2} + \frac{3}{2} \sin 2\theta + 5 \frac{1 + \cos 2\theta}{2}$
$f(\theta) = \frac{1}{2} - \frac{1}{2} \cos 2\theta + \frac{3}{2} \sin 2\theta + \frac{5}{2} + \frac{5}{2} \cos 2\theta$
$f(\theta) = 3 + 2 \cos 2\theta + \frac{3}{2} \sin 2\theta$.
The expression $a \cos x + b \sin x$ ranges between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.
Here,$2 \cos 2\theta + \frac{3}{2} \sin 2\theta$ ranges between $-\sqrt{2^2 + (\frac{3}{2})^2} = -\sqrt{4 + \frac{9}{4}} = -\sqrt{\frac{25}{4}} = -\frac{5}{2}$ and $\frac{5}{2}$.
Thus,the minimum value of $f(\theta) = 3 - \frac{5}{2} = \frac{1}{2}$.
The maximum value of $\frac{1}{f(\theta)}$ is $\frac{1}{1/2} = 2$.

(A) Given $x+y=\frac{\pi}{2}$,we have $y=\frac{\pi}{2}-x$.
Substituting this into the expression $\sin x \cdot \sin y$,we get:
$\sin x \cdot \sin \left(\frac{\pi}{2}-x\right) = \sin x \cdot \cos x$.
Multiplying and dividing by $2$,we get:
$\frac{2 \sin x \cdot \cos x}{2} = \frac{\sin 2x}{2}$.
We know that the range of $\sin 2x$ is $[-1, 1]$.
Therefore,the range of $\frac{\sin 2x}{2}$ is $\left[\frac{-1}{2}, \frac{1}{2}\right]$.
Thus,the maximum value is $\frac{1}{2}$.

(B) Given the function $f(x) = a^{2} \cos^{2} x + b^{2} \sin^{2} x$.
Using the trigonometric identities $\cos^{2} x = \frac{1 + \cos 2x}{2}$ and $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get:
$f(x) = a^{2} \left( \frac{1 + \cos 2x}{2} \right) + b^{2} \left( \frac{1 - \cos 2x}{2} \right)$
$f(x) = \frac{a^{2} + a^{2} \cos 2x + b^{2} - b^{2} \cos 2x}{2}$
$f(x) = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} \right) \cos 2x$.
Since $-1 \leq \cos 2x \leq 1$,the function $f(x)$ attains its minimum value when $\cos 2x = -1$ (given $a^{2} > b^{2}$,so $\frac{a^{2} - b^{2}}{2} > 0$).
Substituting $\cos 2x = -1$:
$f_{\text{min}} = \frac{a^{2} + b^{2}}{2} + \left( \frac{a^{2} - b^{2}}{2} \right) (-1)$
$f_{\text{min}} = \frac{a^{2} + b^{2} - a^{2} + b^{2}}{2} = \frac{2b^{2}}{2} = b^{2}$.
Thus,the minimum value is $b^{2}$.

(B) The range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
For the equation $7 \cos x + 5 \sin x = 2k + 1$,the range of the left side is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.
Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.
Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.
Dividing by $2$: $-4.801 \leq k \leq 3.801$.
The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
Counting these values,we get $8$ integral values.

(D) Given $\alpha+\beta+\gamma=\pi$,so $\gamma = \pi - (\alpha+\beta)$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:
$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin(\beta+\gamma)\sin(\beta-\gamma)$
Since $\beta+\gamma = \pi - \alpha$,$\sin(\beta+\gamma) = \sin(\pi-\alpha) = \sin \alpha$.
Substituting this:
$= \sin^2 \alpha + \sin \alpha \sin(\beta-\gamma)$
$= \sin \alpha [\sin \alpha + \sin(\beta-\gamma)]$
$= \sin \alpha [\sin(\beta+\gamma) + \sin(\beta-\gamma)]$
Using the sum-to-product formula $\sin(x+y) + \sin(x-y) = 2 \sin x \cos y$:
$= \sin \alpha [2 \sin \beta \cos \gamma]$
$= 2 \sin \alpha \sin \beta \cos \gamma$.

(C) Given,$\alpha = \beta + \gamma$.
Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$.
Substituting $\beta$ in the first equation: $\alpha = (\frac{\pi}{2} - \alpha) + \gamma$,which implies $\gamma = 2\alpha - \frac{\pi}{2}$.
Alternatively,using $\gamma = \alpha - \beta$:
$\tan \gamma = \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Since $\beta = \frac{\pi}{2} - \alpha$,then $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$.
Substituting this: $\tan \gamma = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha (\frac{1}{\tan \alpha})} = \frac{\tan \alpha - \tan \beta}{1 + 1} = \frac{\tan \alpha - \tan \beta}{2}$.
Therefore,$2 \tan \gamma = \tan \alpha - \tan \beta$,which gives $\tan \alpha = \tan \beta + 2 \tan \gamma$.

(D) Given the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$.
We know that $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting this into the equation: $1 - 2 \sin^2 \theta \cos^2 \theta + \lambda = 0$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$.
Thus,$1 - 2 \left(\frac{\sin^2 2\theta}{4}\right) + \lambda = 0$,which simplifies to $1 - \frac{\sin^2 2\theta}{2} + \lambda = 0$.
Therefore,$\lambda = \frac{\sin^2 2\theta}{2} - 1$.
Since $0 \leq \sin^2 2\theta \leq 1$,we have $0 \leq \frac{\sin^2 2\theta}{2} \leq \frac{1}{2}$.
Subtracting $1$ from all parts,we get $-1 \leq \frac{\sin^2 2\theta}{2} - 1 \leq -\frac{1}{2}$.
Hence,$\lambda \in \left[-1, -\frac{1}{2}\right]$.