Trigonometry Questions in English

(B) Given $\cos A = \frac{3}{4}$.
We use the product-to-sum formula: $2\sin X \cos Y = \sin(X+Y) + \sin(X-Y)$.
Here,$X = \frac{A}{2}$ and $Y = \frac{5A}{2}$.
$32\sin \frac{A}{2}\cos \frac{5A}{2} = 16 \times (2\sin \frac{A}{2}\cos \frac{5A}{2})$
$= 16 \times [\sin(\frac{A}{2} + \frac{5A}{2}) + \sin(\frac{A}{2} - \frac{5A}{2})]$
$= 16 \times [\sin(3A) + \sin(-2A)]$
$= 16 \times [\sin(3A) - \sin(2A)]$
Using $\sin(3A) = 3\sin A - 4\sin^3 A$ and $\sin(2A) = 2\sin A \cos A$:
$= 16 \times [3\sin A - 4\sin^3 A - 2\sin A \cos A]$
$= 16\sin A [3 - 4\sin^2 A - 2\cos A]$
Since $\cos A = \frac{3}{4}$,$\sin^2 A = 1 - (\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$,so $\sin A = \frac{\sqrt{7}}{4}$.
$= 16 \times \frac{\sqrt{7}}{4} [3 - 4(\frac{7}{16}) - 2(\frac{3}{4})]$
$= 4\sqrt{7} [3 - \frac{7}{4} - \frac{3}{2}]$
$= 4\sqrt{7} [\frac{12 - 7 - 6}{4}] = 4\sqrt{7} [-\frac{1}{4}] = -\sqrt{7}$.

(D) Given,$\tan \theta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}}$.
Since $\theta$ is in the $1^{st}$ quadrant,$\sin \theta = \frac{1}{\sqrt{50}}$ and $\cos \theta = \frac{7}{\sqrt{50}}$.
Since $\phi$ is in the $1^{st}$ quadrant,$\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Now,calculate $\cos 2\phi$ and $\sin 2\phi$:
$\cos 2\phi = 2\cos^2 \phi - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.
$\sin 2\phi = 2\sin \phi \cos \phi = 2(\frac{1}{\sqrt{10}})(\frac{3}{\sqrt{10}}) = \frac{6}{10} = \frac{3}{5}$.
Now,calculate $\cos(\theta + 2\phi)$:
$\cos(\theta + 2\phi) = \cos \theta \cos 2\phi - \sin \theta \sin 2\phi$
$= (\frac{7}{\sqrt{50}})(\frac{8}{10}) - (\frac{1}{\sqrt{50}})(\frac{6}{10})$
$= \frac{56 - 6}{10\sqrt{50}} = \frac{50}{10\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\cos(\theta + 2\phi) = \frac{1}{\sqrt{2}}$,we have $\theta + 2\phi = 45^\circ$.

(D) We have $\frac{\cos A}{1 - \sin A}$.
Multiply the numerator and denominator by $(1 + \sin A)$:
$\frac{\cos A(1 + \sin A)}{(1 - \sin A)(1 + \sin A)} = \frac{\cos A(1 + \sin A)}{1 - \sin^2 A} = \frac{\cos A(1 + \sin A)}{\cos^2 A} = \frac{1 + \sin A}{\cos A}$.
Using half-angle identities $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$ and $\cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}$:
$= \frac{(\cos \frac{A}{2} + \sin \frac{A}{2})^2}{(\cos \frac{A}{2} - \sin \frac{A}{2})(\cos \frac{A}{2} + \sin \frac{A}{2})} = \frac{\cos \frac{A}{2} + \sin \frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}$.
Divide numerator and denominator by $\cos \frac{A}{2}$:
$= \frac{1 + \tan \frac{A}{2}}{1 - \tan \frac{A}{2}} = \tan \left( \frac{\pi}{4} + \frac{A}{2} \right)$.

(A) Given $\sin \alpha = -\frac{3}{5}$ and $\pi < \alpha < \frac{3\pi}{2}$ (i.e.,$\alpha$ is in the $III^{rd}$ quadrant).
In the $III^{rd}$ quadrant,$\cos \alpha$ is negative.
$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - (-\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Since $\pi < \alpha < \frac{3\pi}{2},$ dividing by $2$ gives $\frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4}$.
This implies $\frac{\alpha}{2}$ lies in the $II^{nd}$ quadrant,where $\cos$ is negative.
The half-angle formula is $\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + \cos \alpha}{2}}$.
Substituting the value of $\cos \alpha$:
$\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-4/5)}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(B) Given expression: $\sec 2x - \tan 2x$
$= \frac{1}{\cos 2x} - \frac{\sin 2x}{\cos 2x} = \frac{1 - \sin 2x}{\cos 2x}$
Using the identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,and $1 = \cos^2 x + \sin^2 x$:
$= \frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{(\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$
$= \frac{\cos x - \sin x}{\cos x + \sin x}$
Dividing numerator and denominator by $\cos x$:
$= \frac{1 - \tan x}{1 + \tan x}$
Since $\tan \frac{\pi}{4} = 1$,we can write this as:
$= \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \tan \left( \frac{\pi}{4} - x \right)$.

(B) Given $\sin \theta + \cos \theta = x.$
Squaring both sides,we get $\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = x^2.$
Since $\sin^2 \theta + \cos^2 \theta = 1,$ we have $1 + \sin 2\theta = x^2,$ which implies $\sin 2\theta = x^2 - 1.$
Since the range of $\sin 2\theta$ is $[-1, 1],$ we must have $-1 \le x^2 - 1 \le 1,$ which simplifies to $0 \le x^2 \le 2.$
Now,consider the expression ${\sin ^6}\theta + {\cos ^6}\theta.$
Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b),$ where $a = \sin^2 \theta$ and $b = \cos^2 \theta,$
${\sin ^6}\theta + {\cos ^6}\theta = (\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta).$
$= 1^3 - 3(\sin \theta \cos \theta)^2 (1) = 1 - 3(\frac{\sin 2\theta}{2})^2 = 1 - \frac{3}{4} \sin^2 2\theta.$
Substituting $\sin 2\theta = x^2 - 1,$
$= 1 - \frac{3}{4}(x^2 - 1)^2 = \frac{1}{4}[4 - 3(x^2 - 1)^2].$
This identity holds true only when $\sin 2\theta$ is defined,which requires $x^2 \le 2.$

(A) Given $\tan \theta = t.$
We know the double angle formulas:
$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2t}{1 - t^2}$
$\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - t^2}{1 + t^2}$
Therefore,$\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + t^2}{1 - t^2}$
Now,$\tan 2\theta + \sec 2\theta = \frac{2t}{1 - t^2} + \frac{1 + t^2}{1 - t^2}$
$= \frac{2t + 1 + t^2}{1 - t^2} = \frac{(1 + t)^2}{(1 - t)(1 + t)}$
$= \frac{1 + t}{1 - t}$.

(C) Given expression: $\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }}$
$= \frac{{\sqrt 2 - \sqrt 2 \left( {\frac{1}{{\sqrt 2 }}\sin \alpha + \frac{1}{{\sqrt 2 }}\cos \alpha } \right)}}{{\sqrt 2 \left( {\frac{1}{{\sqrt 2 }}\sin \alpha - \frac{1}{{\sqrt 2 }}\cos \alpha } \right)}}$
$= \frac{{\sqrt 2 - \sqrt 2 \cos \left( {\alpha - \frac{\pi }{4}} \right)}}{{\sqrt 2 \sin \left( {\alpha - \frac{\pi }{4}} \right)}}$
Let $\theta = \alpha - \frac{\pi }{4}$. Then the expression becomes $\frac{{\sqrt 2 (1 - \cos \theta )}}{{\sqrt 2 \sin \theta }} = \frac{{1 - \cos \theta }}{{\sin \theta }}$
Using half-angle identities $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$:
$= \frac{{2{{\sin }^2}(\theta /2)}}{{2\sin (\theta /2)\cos (\theta /2)}} = \tan \frac{\theta }{2}$
Substituting $\theta = \alpha - \frac{\pi }{4}$ back:
$= \tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$.

(C) We know the trigonometric identity: $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$.
Given $\cos \theta = \frac{1}{2}\left( a + \frac{1}{a} \right)$.
Substituting the value of $\cos \theta$ in the identity:
$\cos 3\theta = 4 \left[ \frac{1}{2}\left( a + \frac{1}{a} \right) \right]^3 - 3 \left[ \frac{1}{2}\left( a + \frac{1}{a} \right) \right]$
$\cos 3\theta = 4 \cdot \frac{1}{8} \left( a + \frac{1}{a} \right)^3 - \frac{3}{2} \left( a + \frac{1}{a} \right)$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right)^3 - \frac{3}{2} \left( a + \frac{1}{a} \right)$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left[ \left( a + \frac{1}{a} \right)^2 - 3 \right]$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left[ a^2 + 2 + \frac{1}{a^2} - 3 \right]$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left[ a^2 - 1 + \frac{1}{a^2} \right]$
$\cos 3\theta = \frac{1}{2} \left( a^3 - a + \frac{1}{a} + a - \frac{1}{a} + \frac{1}{a^3} \right)$
$\cos 3\theta = \frac{1}{2} \left( a^3 + \frac{1}{a^3} \right)$.

(B) Given the quadratic equation $25\cos^2\theta + 5\cos\theta - 12 = 0$. Let $x = \cos\theta$,then $25x^2 + 5x - 12 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-5 \pm \sqrt{25 - 4(25)(-12)}}{50} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm \sqrt{1225}}{50} = \frac{-5 \pm 35}{50}$.
This gives two possible values for $\cos\alpha$: $x_1 = \frac{30}{50} = 3/5$ and $x_2 = \frac{-40}{50} = -4/5$.
Since $\pi/2 < \alpha < \pi$,$\alpha$ lies in the second quadrant,where $\cos\alpha$ must be negative. Therefore,$\cos\alpha = -4/5$.
Now,find $\sin\alpha$: $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (-4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$ (positive in the second quadrant).
Finally,$\sin 2\alpha = 2\sin\alpha\cos\alpha = 2(3/5)(-4/5) = -24/25$.

(C) Given $A = 133^\circ$,so $\frac{A}{2} = 66.5^\circ$.
Since $45^\circ < \frac{A}{2} < 90^\circ$,we have $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$.
We know that $1 + \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} + 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} + \cos \frac{A}{2})^2$.
Thus,$\sqrt{1 + \sin A} = |\sin \frac{A}{2} + \cos \frac{A}{2}| = \sin \frac{A}{2} + \cos \frac{A}{2}$ (since both are positive).
Similarly,$1 - \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} - 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} - \cos \frac{A}{2})^2$.
Thus,$\sqrt{1 - \sin A} = |\sin \frac{A}{2} - \cos \frac{A}{2}| = \sin \frac{A}{2} - \cos \frac{A}{2}$ (since $\sin \frac{A}{2} > \cos \frac{A}{2}$).
Subtracting the two expressions: $\sqrt{1 + \sin A} - \sqrt{1 - \sin A} = (\sin \frac{A}{2} + \cos \frac{A}{2}) - (\sin \frac{A}{2} - \cos \frac{A}{2}) = 2\cos \frac{A}{2}$.

(D) Given $\sin A = \frac{4}{5}$ and $90^\circ < A < 180^\circ$ (Second Quadrant).
Since $\sin A$ is positive and $\cos A$ is negative in the second quadrant,$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (4/5)^2} = -\sqrt{1 - 16/25} = -\sqrt{9/25} = -3/5$.
Now,$\tan A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -4/3$.
Using the half-angle formula: $\tan A = \frac{2 \tan(A/2)}{1 - \tan^2(A/2)}$.
Let $P = \tan(A/2)$. Then $-4/3 = \frac{2P}{1 - P^2}$.
$-4(1 - P^2) = 6P \implies -4 + 4P^2 = 6P \implies 4P^2 - 6P - 4 = 0$.
Dividing by $2$: $2P^2 - 3P - 2 = 0$.
Factoring the quadratic: $(2P + 1)(P - 2) = 0$.
So,$P = -1/2$ or $P = 2$.
Since $90^\circ < A < 180^\circ$,then $45^\circ < A/2 < 90^\circ$. In this range,$\tan(A/2)$ must be positive.
Therefore,$\tan(A/2) = 2$.

(B) Given: $2\tan A = 3\tan B \implies \tan A = \frac{3}{2}\tan B$.
Let $\tan B = t$,then $\tan A = \frac{3}{2}t$.
We know that $\sin 2B = \frac{2t}{1 + t^2}$ and $\cos 2B = \frac{1 - t^2}{1 + t^2}$.
Substituting these into the expression $\frac{\sin 2B}{5 - \cos 2B}$:
$= \frac{\frac{2t}{1 + t^2}}{5 - \frac{1 - t^2}{1 + t^2}} = \frac{2t}{5(1 + t^2) - (1 - t^2)} = \frac{2t}{5 + 5t^2 - 1 + t^2} = \frac{2t}{4 + 6t^2} = \frac{t}{2 + 3t^2}$.
Now,consider $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{3}{2}t - t}{1 + (\frac{3}{2}t)(t)} = \frac{\frac{1}{2}t}{1 + \frac{3}{2}t^2} = \frac{t}{2 + 3t^2}$.
Since both expressions are equal,the result is $\tan(A - B)$.

(B) Given the equation: $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$.
Using the expansion formulas $\cos(A - B) = \cos A \cos B + \sin A \sin B$ and $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:
$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \left( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right)$.
Expanding the right side:
$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$.
Rearranging the terms to group $\sin$ and $\cos$ products:
$3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$.
Dividing both sides by $3 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$,we obtain:
$\frac{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{1}{3}$.
Therefore,$\tan \frac{\alpha}{2} \tan \frac{\beta}{2} = \frac{1}{3}$.

(B) Given,$\sqrt x + \frac{1}{{\sqrt x }} = 2\cos \theta$ --- $(i)$
Squaring both sides,we get:
$x + \frac{1}{x} + 2 = 4\cos^2 \theta$
$x + \frac{1}{x} = 4\cos^2 \theta - 2 = 2(2\cos^2 \theta - 1) = 2\cos 2\theta$ --- $(ii)$
Squaring both sides of $(ii)$:
$x^2 + \frac{1}{x^2} + 2 = 4\cos^2 2\theta$
$x^2 + \frac{1}{x^2} = 4\cos^2 2\theta - 2 = 2(2\cos^2 2\theta - 1) = 2\cos 4\theta$ --- $(iii)$
Now,cubing both sides of $(iii)$:
$(x^2 + \frac{1}{x^2})^3 = (2\cos 4\theta)^3$
$x^6 + \frac{1}{x^6} + 3(x^2 \cdot \frac{1}{x^2})(x^2 + \frac{1}{x^2}) = 8\cos^3 4\theta$
$x^6 + \frac{1}{x^6} + 3(2\cos 4\theta) = 8\cos^3 4\theta$
$x^6 + \frac{1}{x^6} = 8\cos^3 4\theta - 6\cos 4\theta$
$x^6 + \frac{1}{x^6} = 2(4\cos^3 4\theta - 3\cos 4\theta)$
Using the identity $\cos 3A = 4\cos^3 A - 3\cos A$,we get:
$x^6 + \frac{1}{x^6} = 2\cos(3 \cdot 4\theta) = 2\cos 12\theta$.

(B) Given equations are:
$\sin 2\theta + \sin 2\phi = 1/2$ --- $(i)$
$\cos 2\theta + \cos 2\phi = 3/2$ --- $(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$(\sin 2\theta + \sin 2\phi)^2 + (\cos 2\theta + \cos 2\phi)^2 = (1/2)^2 + (3/2)^2$
$(\sin^2 2\theta + \cos^2 2\theta) + (\sin^2 2\phi + \cos^2 2\phi) + 2(\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi) = 1/4 + 9/4$
$1 + 1 + 2 \cos(2\theta - 2\phi) = 10/4$
$2 + 2 \cos(2\theta - 2\phi) = 5/2$
$2 \cos(2\theta - 2\phi) = 5/2 - 2 = 1/2$
$\cos(2\theta - 2\phi) = 1/4$
Using the identity $\cos 2A = 2 \cos^2 A - 1$:
$2 \cos^2(\theta - \phi) - 1 = 1/4$
$2 \cos^2(\theta - \phi) = 1 + 1/4 = 5/4$
$\cos^2(\theta - \phi) = 5/8$.

(A) Given expression: $E = \cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2\sin^2 \phi$
Using the identity $2\sin \theta \sin \phi = \cos(\theta - \phi) - \cos(\theta + \phi)$:
$E = \cos 2(\theta + \phi) - 2\cos(\theta + \phi)[\cos(\theta - \phi) - \cos(\theta + \phi)] + 2\sin^2 \phi$
$E = \cos 2(\theta + \phi) - 2\cos(\theta + \phi)\cos(\theta - \phi) + 2\cos^2(\theta + \phi) + 2\sin^2 \phi$
Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \cos 2(\theta + \phi) - [\cos 2\theta + \cos 2\phi] + 2\cos^2(\theta + \phi) + 2\sin^2 \phi$
Using $2\cos^2 A = 1 + \cos 2A$:
$E = \cos 2(\theta + \phi) - \cos 2\theta - \cos 2\phi + 1 + \cos 2(\theta + \phi) + 2\sin^2 \phi$
Since $1 - \cos 2\phi = 2\sin^2 \phi$,the expression simplifies to:
$E = 2\cos 2(\theta + \phi) - \cos 2\theta + 2\sin^2 \phi + 2\sin^2 \phi - \cos 2\phi$
Alternatively,testing with $\theta = \phi = \frac{\pi}{4}$:
$E = \cos 2(\frac{\pi}{2}) - 4\cos(\frac{\pi}{2})\sin(\frac{\pi}{4})\sin(\frac{\pi}{4}) + 2\sin^2(\frac{\pi}{4}) = \cos \pi - 0 + 2(\frac{1}{2}) = -1 + 1 = 0$.
Checking options for $\theta = \frac{\pi}{4}$:
$(a) \cos 2(\frac{\pi}{4}) = \cos \frac{\pi}{2} = 0$.
Thus,option $(a)$ is correct.

(C) We evaluate each option:
$A) \sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$,which is irrational.
$B) \cos 15^\circ = \cos (45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$,which is irrational.
$C) \sin 15^\circ \cos 15^\circ = \frac{1}{2} (2 \sin 15^\circ \cos 15^\circ) = \frac{1}{2} \sin (2 \times 15^\circ) = \frac{1}{2} \sin 30^\circ = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$,which is rational.
$D) \sin 15^\circ \cos 75^\circ = \sin 15^\circ \sin (90^\circ - 75^\circ) = \sin 15^\circ \sin 15^\circ = \sin^2 15^\circ = \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right)^2 = \frac{3 + 1 - 2\sqrt{3}}{8} = \frac{4 - 2\sqrt{3}}{8} = \frac{2 - \sqrt{3}}{4}$,which is irrational.
Thus,the correct option is $C$.

(B) Given: $\sin A + \cos A = \sqrt{2}$.
Squaring both sides:
$(\sin A + \cos A)^2 = (\sqrt{2})^2$
$\sin^2 A + \cos^2 A + 2 \sin A \cos A = 2$
Since $\sin^2 A + \cos^2 A = 1$,we have:
$1 + 2 \sin A \cos A = 2$
$2 \sin A \cos A = 1$
$\sin 2A = 1 = \sin 90^\circ$
$2A = 90^\circ \implies A = 45^\circ$.
Now,calculate $\cos^2 A$:
$\cos^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.

(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$
Take $2$ as a common factor: $2(\cos^2 \theta - \sin^2 \theta) = 1$
Using the trigonometric identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we get:
$2\cos 2\theta = 1$
$\cos 2\theta = \frac{1}{2}$
Since $\cos 60^\circ = \frac{1}{2}$,we have:
$2\theta = 60^\circ$
$\theta = 30^\circ$

(C) Given $\sin \alpha = \frac{336}{625}$ and $450^\circ < \alpha < 540^\circ$.
Since $450^\circ < \alpha < 540^\circ$,$\alpha$ lies in the second quadrant $(90^\circ < \alpha - 360^\circ < 180^\circ)$.
In the second quadrant,$\cos \alpha$ is negative.
$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left( \frac{336}{625} \right)^2} = -\sqrt{\frac{625^2 - 336^2}{625^2}} = -\sqrt{\frac{(625-336)(625+336)}{625^2}} = -\sqrt{\frac{289 \times 961}{625^2}} = -\frac{17 \times 31}{625} = -\frac{527}{625}$.
Now,$225^\circ < \frac{\alpha}{2} < 270^\circ$,so $\frac{\alpha}{2}$ is in the third quadrant.
$\cos \left( \frac{\alpha}{2} \right) = -\sqrt{\frac{1 + \cos \alpha}{2}} = -\sqrt{\frac{1 - \frac{527}{625}}{2}} = -\sqrt{\frac{98}{1250}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.
Finally,$112.5^\circ < \frac{\alpha}{4} < 135^\circ$,so $\frac{\alpha}{4}$ is in the second quadrant.
In the second quadrant,$\sin \left( \frac{\alpha}{4} \right)$ is positive.
$\sin \left( \frac{\alpha}{4} \right) = \sqrt{\frac{1 - \cos(\alpha/2)}{2}} = \sqrt{\frac{1 - (-7/25)}{2}} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(B) Given: $\tan^2 \theta = 2\tan^2 \phi + 1$
Adding $1$ to both sides: $1 + \tan^2 \theta = 2\tan^2 \phi + 2$
$\Rightarrow \sec^2 \theta = 2(1 + \tan^2 \phi)$
$\Rightarrow \sec^2 \theta = 2\sec^2 \phi$
Taking the reciprocal: $\cos^2 \theta = \frac{1}{2} \cos^2 \phi$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$\frac{1 + \cos 2\theta}{2} = \frac{1}{2} \cos^2 \phi$
$1 + \cos 2\theta = \cos^2 \phi$
$1 + \cos 2\theta = 1 - \sin^2 \phi$
$\cos 2\theta + \sin^2 \phi = 0$
Alternatively,let $\theta = 45^\circ$. Then $\tan^2 45^\circ = 2\tan^2 \phi + 1 \Rightarrow 1 = 2\tan^2 \phi + 1 \Rightarrow \tan^2 \phi = 0 \Rightarrow \phi = 0^\circ$.
Thus,$\cos(2 \times 45^\circ) + \sin^2 0^\circ = \cos 90^\circ + 0 = 0 + 0 = 0$.

(C) Given expression: $E = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8}$
Since $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{7\pi}{8} = \cos(\pi - \frac{\pi}{8}) = -\cos \frac{\pi}{8}$ and $\cos \frac{5\pi}{8} = \cos(\pi - \frac{3\pi}{8}) = -\cos \frac{3\pi}{8}$.
Raising these to the power of $4$,we get $\cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}$ and $\cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}$.
Substituting these into the expression: $E = 2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8})$.
Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,where $a = \cos^2 \frac{\pi}{8}$ and $b = \cos^2 \frac{3\pi}{8}$:
$E = 2[(\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8})^2 - 2\cos^2 \frac{\pi}{8} \cos^2 \frac{3\pi}{8}]$
Note that $\cos^2 \frac{3\pi}{8} = \sin^2 \frac{\pi}{8}$,so $\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} = \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1$.
$E = 2[1^2 - 2(\cos \frac{\pi}{8} \cos \frac{3\pi}{8})^2] = 2[1 - 2(\frac{1}{2} (2 \cos \frac{\pi}{8} \cos \frac{3\pi}{8}))^2]$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$2 \cos \frac{\pi}{8} \cos \frac{3\pi}{8} = \cos(\frac{4\pi}{8}) + \cos(-\frac{2\pi}{8}) = \cos \frac{\pi}{2} + \cos \frac{\pi}{4} = 0 + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
$E = 2[1 - 2(\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2] = 2[1 - 2(\frac{1}{2\sqrt{2}})^2] = 2[1 - 2(\frac{1}{8})] = 2[1 - \frac{1}{4}] = 2(\frac{3}{4}) = \frac{3}{2}$.

(D) Given: $\sin x + \cos x = \frac{1}{5}$
Squaring both sides:
$(\sin x + \cos x)^2 = (\frac{1}{5})^2$
$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}{25}$
Since $\sin^2 x + \cos^2 x = 1$ and $2 \sin x \cos x = \sin 2x$:
$1 + \sin 2x = \frac{1}{25}$
$\sin 2x = \frac{1}{25} - 1 = -\frac{24}{25}$
Now,$\cos^2 2x = 1 - \sin^2 2x = 1 - (-\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$
$\cos 2x = \pm \frac{7}{25}$
Since $\sin 2x$ is negative,$2x$ lies in the $III$ or $IV$ quadrant. If we assume the standard identity approach:
$\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{-24/25}{\pm 7/25} = \mp \frac{24}{7}$.
Given the options,the magnitude is $\frac{24}{7}$.

(C) Given expression: $\cos^2 A(3 - 4\cos^2 A)^2 + \sin^2 A(3 - 4\sin^2 A)^2$
$= (3\cos A - 4\cos^3 A)^2 + (3\sin A - 4\sin^3 A)^2$
Using the triple angle identities: $\cos 3A = 4\cos^3 A - 3\cos A$ and $\sin 3A = 3\sin A - 4\sin^3 A$.
Note that $(3\cos A - 4\cos^3 A)^2 = (-(4\cos^3 A - 3\cos A))^2 = (-\cos 3A)^2 = \cos^2 3A$.
So,the expression becomes $\cos^2 3A + \sin^2 3A$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $\cos^2 3A + \sin^2 3A = 1$.
Trick: Substitute $A = 0^\circ$. The expression becomes $\cos^2 0(3 - 4\cos^2 0)^2 + \sin^2 0(3 - 4\sin^2 0)^2 = 1(3 - 4)^2 + 0 = (-1)^2 = 1$.

(C) We know that $1 = \sec^2 A - \tan^2 A$. Substituting this in the numerator:
$\frac{{\tan A + \sec A - (\sec^2 A - \tan^2 A)}}{{\tan A - \sec A + 1}}$
$= \frac{{(\tan A + \sec A) - (\sec A - \tan A)(\sec A + \tan A)}}{{\tan A - \sec A + 1}}$
$= \frac{{(\tan A + \sec A)(1 - (\sec A - \tan A))}}{{\tan A - \sec A + 1}}$
$= \frac{{(\tan A + \sec A)(1 - \sec A + \tan A)}}{{\tan A - \sec A + 1}}$
Since the term $(1 - \sec A + \tan A)$ is common in the numerator and denominator,they cancel out:
$= \tan A + \sec A$
$= \frac{{\sin A}}{{\cos A}} + \frac{1}{{\cos A}}$
$= \frac{{1 + \sin A}}{{\cos A}}$
Thus,the correct option is $C$.

(D) We know that $1 - \sin A = 1 - \cos \left( {\frac{\pi }{2} - A} \right) = 2\sin^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$ and $1 + \sin A = 1 + \cos \left( {\frac{\pi }{2} - A} \right) = 2\cos^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$.
Substituting these into the expression:
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = \sqrt {\frac{{2\sin^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}{{2\cos^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}} = \sqrt {\tan^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)} = \tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$.

(A) Let the given expression be $E = \frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1$.
Using the identities $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$ and $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$,we have:
$N = \sin 3\theta - \cos 3\theta = (3\sin \theta - 4\sin^3 \theta) - (4\cos^3 \theta - 3\cos \theta)$
$N = 3(\sin \theta + \cos \theta) - 4(\sin^3 \theta + \cos^3 \theta)$
Using the sum of cubes formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$:
$N = 3(\sin \theta + \cos \theta) - 4(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$N = (\sin \theta + \cos \theta) [3 - 4(1 - \sin \theta \cos \theta)]$
Now,divide by $D = \sin \theta + \cos \theta$:
$\frac{N}{D} = 3 - 4 + 4\sin \theta \cos \theta = 4\sin \theta \cos \theta - 1$
Adding $1$ to the expression:
$E = (4\sin \theta \cos \theta - 1) + 1 = 4\sin \theta \cos \theta$
Using the double angle identity $2\sin \theta \cos \theta = \sin 2\theta$:
$E = 2(2\sin \theta \cos \theta) = 2\sin 2\theta$.

(B) We use the identity $\tan A = \frac{\sin 2A}{1 + \cos 2A}$.
Setting $A = 7\frac{1}{2}^\circ$,we get $2A = 15^\circ$.
Thus,$\tan 7\frac{1}{2}^\circ = \frac{\sin 15^\circ}{1 + \cos 15^\circ}$.
We know that $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Substituting these values: $\tan 7\frac{1}{2}^\circ = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{1 + \frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{4 + \sqrt{6} + \sqrt{2}}$.
Multiplying the numerator and denominator by the conjugate $(4 + \sqrt{2}) - \sqrt{6}$,we simplify to get $\sqrt{6} - \sqrt{3} + \sqrt{2} - 2$.

(B) Given $\sin \frac{\theta}{2} = \sqrt{\frac{x - 1}{2x}}$.
We know that $\cos \theta = 1 - 2\sin^2 \frac{\theta}{2} = 1 - 2\left(\frac{x - 1}{2x}\right) = 1 - \frac{x - 1}{x} = \frac{x - x + 1}{x} = \frac{1}{x}$.
Also,$\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
Since $\sin \frac{\theta}{2} = \sqrt{\frac{x - 1}{2x}}$,then $\cos \frac{\theta}{2} = \sqrt{1 - \sin^2 \frac{\theta}{2}} = \sqrt{1 - \frac{x - 1}{2x}} = \sqrt{\frac{2x - x + 1}{2x}} = \sqrt{\frac{x + 1}{2x}}$.
Thus,$\sin \theta = 2 \sqrt{\frac{x - 1}{2x}} \cdot \sqrt{\frac{x + 1}{2x}} = 2 \frac{\sqrt{x^2 - 1}}{2x} = \frac{\sqrt{x^2 - 1}}{x}$.
Finally,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1}$.

(C) The expression is of the form $a\cos \theta + b\sin \theta$,where $a = 3$ and $b = -4$.
The maximum value of the expression $a\cos \theta + b\sin \theta$ is given by the formula $\sqrt{a^2 + b^2}$.
Substituting the values of $a$ and $b$:
Maximum value $= \sqrt{(3)^2 + (-4)^2}$
$= \sqrt{9 + 16}$
$= \sqrt{25}$
$= 5$.
Thus,the maximum value of $3\cos \theta - 4\sin \theta$ is $5$.

(D) Let $f(\theta) = 5 \sin^2 \theta + 4 \cos^2 \theta$.
We know that $\sin^2 \theta + \cos^2 \theta = 1$,so $\cos^2 \theta = 1 - \sin^2 \theta$.
Substituting this into the expression:
$f(\theta) = 5 \sin^2 \theta + 4(1 - \sin^2 \theta)$
$f(\theta) = 5 \sin^2 \theta + 4 - 4 \sin^2 \theta$
$f(\theta) = \sin^2 \theta + 4$.
Since the minimum value of $\sin^2 \theta$ is $0$,the minimum value of $f(\theta)$ is $0 + 4 = 4$.

(C) Given expression: $f(x) = \cos^2 \left( \frac{\pi}{3} - x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)$.
Using the identity $\cos^2 A - \cos^2 B = \sin(B - A) \sin(B + A)$:
Let $A = \frac{\pi}{3} - x$ and $B = \frac{\pi}{3} + x$.
Then $B - A = (\frac{\pi}{3} + x) - (\frac{\pi}{3} - x) = 2x$.
And $B + A = (\frac{\pi}{3} + x) + (\frac{\pi}{3} - x) = \frac{2\pi}{3}$.
Thus,$f(x) = \sin(2x) \sin\left( \frac{2\pi}{3} \right)$.
Since $\sin\left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}$,we have $f(x) = \frac{\sqrt{3}}{2} \sin(2x)$.
The maximum value of $\sin(2x)$ is $1$.
Therefore,the maximum value of the expression is $\frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2}$.

(A) We know that for any real number $x \neq 0$,the square of a real number is always non-negative,so $(x - \frac{1}{x})^2 \ge 0$.
Expanding this,we get $x^2 + \frac{1}{x^2} - 2 \ge 0$,which implies $x^2 + \frac{1}{x^2} \ge 2$.
Let $x = \tan \theta$. Since $\cot \theta = \frac{1}{\tan \theta}$,substituting this into the inequality gives $\tan^2 \theta + \cot^2 \theta \ge 2$.
Thus,the expression is always greater than or equal to $2$.

(A) Let $f(x) = \sqrt{3} \cos x + \sin x$.
We can rewrite this expression by multiplying and dividing by $2$:
$f(x) = 2 \left( \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \right)$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$,we get:
$f(x) = 2 \sin(x + 60^\circ)$.
The maximum value of the sine function is $1$,which occurs when the angle is $90^\circ$.
Therefore,for the maximum value of $f(x)$:
$x + 60^\circ = 90^\circ$
$x = 90^\circ - 60^\circ = 30^\circ$.

(D) The expression is of the form $a \cos x + b \sin x + c$.
We know that the range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$.
Therefore,the range of $3 \cos x + 4 \sin x$ is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$.
The minimum value of $3 \cos x + 4 \sin x$ is $-5$.
Thus,the minimum value of $3 \cos x + 4 \sin x + 5$ is $-5 + 5 = 0$.

(B) Let $f(x) = \sin x \cos x$.
Multiplying and dividing by $2$,we get $f(x) = \frac{2 \sin x \cos x}{2} = \frac{1}{2} \sin(2x)$.
We know that the range of the sine function is $[-1, 1]$,so $-1 \le \sin(2x) \le 1$.
Multiplying the entire inequality by $\frac{1}{2}$,we get $-\frac{1}{2} \le \frac{1}{2} \sin(2x) \le \frac{1}{2}$.
Therefore,the greatest value is $\frac{1}{2}$ and the least value is $-\frac{1}{2}$.

(B) Let $f(\theta) = \cos \theta + \sin \theta$.
We can rewrite this expression by multiplying and dividing by $\sqrt{2}$:
$f(\theta) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta \right)$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get:
$f(\theta) = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right)$
Since the range of the sine function is $[-1, 1]$,we have:
$-1 \le \sin \left( \theta + \frac{\pi}{4} \right) \le 1$
Multiplying by $\sqrt{2}$,we get:
$-\sqrt{2} \le \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \le \sqrt{2}$
Therefore,the minimum value of the expression is $-\sqrt{2}$.

(B) Let $f(x) = 4\sin^2 x + 3\cos^2 x$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we can write $\cos^2 x = 1 - \sin^2 x$.
Substituting this into the expression: $f(x) = 4\sin^2 x + 3(1 - \sin^2 x)$.
$f(x) = 4\sin^2 x + 3 - 3\sin^2 x$.
$f(x) = \sin^2 x + 3$.
Since the range of $\sin x$ is $[-1, 1]$,the range of $\sin^2 x$ is $[0, 1]$.
To find the maximum value,we take the maximum value of $\sin^2 x$,which is $1$.
Maximum value $= 1 + 3 = 4$.

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.
We can rewrite this expression as $f(x) = \sqrt{2} \left[ \frac{1}{\sqrt{2}} \sin \left( x + \frac{\pi}{6} \right) + \frac{1}{\sqrt{2}} \cos \left( x + \frac{\pi}{6} \right) \right]$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{6} + \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} \right)$.
The maximum value of the sine function is $1$,which occurs when the argument is $\frac{\pi}{2}$.
Setting $x + \frac{5\pi}{12} = \frac{\pi}{2}$,we get $x = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi - 5\pi}{12} = \frac{\pi}{12}$.
Since $\frac{\pi}{12}$ lies within the interval $\left( 0, \frac{\pi}{2} \right)$,the maximum value is attained at $x = \frac{\pi}{12}$.

(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states that $\frac{a+b}{2} \ge \sqrt{ab}$.
Let $a = 9\tan^2\theta$ and $b = 4\cot^2\theta$.
Then,$\frac{9\tan^2\theta + 4\cot^2\theta}{2} \ge \sqrt{9\tan^2\theta \cdot 4\cot^2\theta}$.
Since $\tan\theta \cdot \cot\theta = 1$,we have $\sqrt{9 \cdot 4 \cdot 1} = \sqrt{36} = 6$.
Therefore,$\frac{9\tan^2\theta + 4\cot^2\theta}{2} \ge 6$.
Multiplying both sides by $2$,we get $9\tan^2\theta + 4\cot^2\theta \ge 12$.
Thus,the minimum value is $12$.

(C) Given that $\alpha + \beta + \gamma = \pi$.
For a triangle with angles $\alpha, \beta, \gamma$,the sum of the sines of the angles is given by $\sin \alpha + \sin \beta + \sin \gamma = 4 \cos(\alpha/2) \cos(\beta/2) \cos(\gamma/2)$.
Since $\alpha, \beta, \gamma$ are angles of a triangle (or satisfy the sum condition),they must be in the interval $(0, \pi)$.
Consequently,$\alpha/2, \beta/2, \gamma/2$ are in the interval $(0, \pi/2)$.
In the interval $(0, \pi/2)$,the cosine function is always positive.
Therefore,$4 \cos(\alpha/2) \cos(\beta/2) \cos(\gamma/2) > 0$.
Thus,the expression $\sin \alpha + \sin \beta + \sin \gamma$ is always positive.

(D) The expression is of the form $a\sin \theta + b\cos \theta$.
The range of this expression is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$.
Therefore,the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$.
The minimum value is $-5$.

(A) To find the maximum value of the expression $f(x) = \sin x - \cos x$,we can rewrite it in the form $R \sin(x - \alpha)$.
We know that any expression of the form $a \sin x + b \cos x$ has a range of $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 1$ and $b = -1$.
The maximum value is given by $\sqrt{a^2 + b^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the maximum value is $\sqrt{2}$.

(D) Given $A = \cos^2 \theta + \sin^4 \theta$.
We can write this as $A = \cos^2 \theta + \sin^2 \theta \cdot \sin^2 \theta$.
Since $\sin^2 \theta \le 1$,we have $A \le \cos^2 \theta + \sin^2 \theta$,which implies $A \le 1$.
Now,express $A$ in terms of $\sin^2 \theta$:
$A = (1 - \sin^2 \theta) + \sin^4 \theta$.
Let $x = \sin^2 \theta$,where $0 \le x \le 1$.
Then $A = x^2 - x + 1$.
Completing the square: $A = (x - 1/2)^2 + 3/4$.
Since $(x - 1/2)^2 \ge 0$,the minimum value of $A$ is $3/4$ (at $x = 1/2$).
Thus,$3/4 \le A \le 1$.

(B) Given $A = \sin^2 \theta + \cos^4 \theta$.
Since $\cos^2 \theta \le 1$,we have $\cos^4 \theta \le \cos^2 \theta$.
Thus,$A = \sin^2 \theta + \cos^4 \theta \le \sin^2 \theta + \cos^2 \theta = 1$. So,$A \le 1$.
Now,express $A$ in terms of $\cos^2 \theta$:
$A = (1 - \cos^2 \theta) + \cos^4 \theta = \cos^4 \theta - \cos^2 \theta + 1$.
Let $x = \cos^2 \theta$,where $0 \le x \le 1$.
Then $A = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.
Since $(x - \frac{1}{2})^2 \ge 0$,the minimum value is $\frac{3}{4}$ (at $x = \frac{1}{2}$).
Since $x \in [0, 1]$,the maximum value occurs at the boundaries $x=0$ or $x=1$,which is $1$.
Therefore,$\frac{3}{4} \le A \le 1$.

(C) The expression is $f(x) = \sqrt{3} \sin x + \cos x$.
We can rewrite this by multiplying and dividing by $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
$f(x) = 2 \left( \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have $\sin(x + 30^\circ) = \sin x \cos 30^\circ + \cos x \sin 30^\circ = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x$.
Thus,$f(x) = 2 \sin(x + 30^\circ)$.
The maximum value of the sine function is $1$,which occurs when the angle is $90^\circ$.
Therefore,$x + 30^\circ = 90^\circ$,which gives $x = 60^\circ$.

(A) Given $\alpha + \beta - \gamma = \pi$,which implies $\gamma = \alpha + \beta - \pi$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A - B) \sin(A + B)$:
$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + (\sin^2 \beta - \sin^2 \gamma)$
$= \sin^2 \alpha + \sin(\beta - \gamma) \sin(\beta + \gamma)$
Since $\beta - \gamma = \pi - \alpha$,we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha$.
Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi$.
Alternatively,using $\gamma = \alpha + \beta - \pi$,then $\sin \gamma = \sin(\alpha + \beta - \pi) = -\sin(\alpha + \beta)$.
So,$\sin^2 \gamma = \sin^2(\alpha + \beta)$.
Expression $= \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta)$
$= \sin^2 \alpha + \sin^2 \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2$
$= \sin^2 \alpha + \sin^2 \beta - (\sin^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta)$
$= \sin^2 \alpha(1 - \cos^2 \beta) + \sin^2 \beta(1 - \cos^2 \alpha) - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$
$= \sin^2 \alpha \sin^2 \beta + \sin^2 \beta \sin^2 \alpha - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$
$= 2 \sin \alpha \sin \beta (\sin \alpha \sin \beta - \cos \alpha \cos \beta)$
$= 2 \sin \alpha \sin \beta (- \cos(\alpha + \beta))$
Since $\alpha + \beta = \pi + \gamma$,$\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma$.
Therefore,the expression $= 2 \sin \alpha \sin \beta (-(- \cos \gamma)) = 2 \sin \alpha \sin \beta \cos \gamma$.

(D) Given that $ABCD$ is a cyclic quadrilateral.
In a cyclic quadrilateral,the sum of opposite angles is $180^\circ$.
Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.
From $A + C = 180^\circ$,we get $A = 180^\circ - C$.
Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$.
This implies $\cos A + \cos C = 0$.
Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.
Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$.
This implies $\cos B + \cos D = 0$.
Adding these two equations,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

(C) Given $A + B + C = \pi$.
Let $S = \frac{\cos A}{\sin B \sin C} + \frac{\cos B}{\sin C \sin A} + \frac{\cos C}{\sin A \sin B}$.
Taking the common denominator $\sin A \sin B \sin C$,we get:
$S = \frac{\cos A \sin A + \cos B \sin B + \cos C \sin C}{\sin A \sin B \sin C}$.
Multiply numerator and denominator by $2$:
$S = \frac{2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C}{2 \sin A \sin B \sin C} = \frac{\sin 2A + \sin 2B + \sin 2C}{2 \sin A \sin B \sin C}$.
Using the identity for a triangle where $A + B + C = \pi$,$\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$.
Substituting this into the expression:
$S = \frac{4 \sin A \sin B \sin C}{2 \sin A \sin B \sin C} = 2$.