Trigonometrical ratios of multiple and sub-multiple angles Questions in English

(A) To evaluate the product $P = \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$:
$P = \frac{1}{2 \sin A} (2 \sin A \cos A) \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$P = \frac{1}{2 \sin A} (\sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A$
Multiply and divide by $2$:
$P = \frac{1}{2^2 \sin A} (2 \sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A = \frac{\sin 4 A}{2^2 \sin A} \cos 4 A \ldots \cos 2^{n-1} A$
Continuing this process $n$ times,we obtain:
$P = \frac{\sin 2^n A}{2^n \sin A}$

(B) Given $\tan \theta = \frac{-3}{4}$. Since $\tan \theta < 0$ and $\theta$ is not in the second quadrant,$\theta$ must be in the fourth quadrant.
In the fourth quadrant,$\sin \theta = \frac{-3}{5}$ and $\cos \theta = \frac{4}{5}$.
Using the half-angle formula,$\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} = \frac{1 - 4/5}{-3/5} = \frac{1/5}{-3/5} = \frac{-1}{3}$.
Also,$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \times (\frac{-3}{5}) \times (\frac{4}{5}) = \frac{-24}{25}$.
Therefore,$\tan \frac{\theta}{2} + \sin 2 \theta = \frac{-1}{3} + (\frac{-24}{25}) = \frac{-25 - 72}{75} = \frac{-97}{75}$.

(B) Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos^3 x = \frac{\cos 3x + 3 \cos x}{4}$.
Applying this to each term:
$\frac{\cos 3 \theta + 3 \cos \theta}{4} + \frac{\cos(2 \pi + 3 \theta) + 3 \cos(\frac{2 \pi}{3} + \theta)}{4} + \frac{\cos(4 \pi + 3 \theta) + 3 \cos(\frac{4 \pi}{3} + \theta)}{4} = \alpha \cos 3 \theta$
Since $\cos(2 \pi + 3 \theta) = \cos 3 \theta$ and $\cos(4 \pi + 3 \theta) = \cos 3 \theta$,the equation becomes:
$\frac{3 \cos 3 \theta + 3 \cos \theta + 3 \cos(\frac{2 \pi}{3} + \theta) + 3 \cos(\frac{4 \pi}{3} + \theta)}{4} = \alpha \cos 3 \theta$
Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$\cos(\frac{2 \pi}{3} + \theta) + \cos(\frac{4 \pi}{3} + \theta) = 2 \cos(\pi + \theta) \cos(-\frac{\pi}{3}) = 2(-\cos \theta)(\frac{1}{2}) = -\cos \theta$
Substituting this back:
$\frac{3 \cos 3 \theta + 3 \cos \theta + 3(-\cos \theta)}{4} = \alpha \cos 3 \theta$
$\frac{3 \cos 3 \theta}{4} = \alpha \cos 3 \theta$
Thus,$\alpha = \frac{3}{4}$.

(B) Given $\frac{\tan 3A}{\tan A} = a$.
We know that $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$.
So,$\frac{3\tan A - \tan^3 A}{\tan A(1 - 3\tan^2 A)} = a$.
$\frac{3 - \tan^2 A}{1 - 3\tan^2 A} = a$.
$3 - \tan^2 A = a - 3a\tan^2 A$.
$\tan^2 A(3a - 1) = a - 3$.
$\tan^2 A = \frac{a - 3}{3a - 1}$.
Now,$\frac{\sin 3A}{\sin A} = \frac{3\sin A - 4\sin^3 A}{\sin A} = 3 - 4\sin^2 A$.
Using $\sin^2 A = \frac{\tan^2 A}{1 + \tan^2 A} = \frac{\frac{a-3}{3a-1}}{1 + \frac{a-3}{3a-1}} = \frac{a-3}{3a-1+a-3} = \frac{a-3}{4a-4} = \frac{a-3}{4(a-1)}$.
Therefore,$\frac{\sin 3A}{\sin A} = 3 - 4\left(\frac{a-3}{4(a-1)}\right) = 3 - \frac{a-3}{a-1} = \frac{3(a-1) - (a-3)}{a-1} = \frac{3a - 3 - a + 3}{a-1} = \frac{2a}{a-1}$.

(D) We know that $(1 + \sec \theta) = \frac{1 + \cos \theta}{\cos \theta} = \frac{2 \cos^2 \frac{\theta}{2}}{\cos \theta}$.
Also,$\tan \frac{\theta}{2} (1 + \sec \theta) = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot \frac{2 \cos^2 \frac{\theta}{2}}{\cos \theta} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Applying this repeatedly,$f_n(x) = \tan \frac{x}{2} (1 + \sec x) (1 + \sec 2x) \dots (1 + \sec 2^n x) = \tan 2^n x$.
For $x = \frac{\pi}{16}$:
$f_2\left(\frac{\pi}{16}\right) = \tan \left(2^2 \cdot \frac{\pi}{16}\right) = \tan \left(4 \cdot \frac{\pi}{16}\right) = \tan \frac{\pi}{4} = 1$.
Thus,option $D$ is correct.

(B) We know that $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$ and $\sin 2x = 2 \sin x \cos x$,we get $\cot x - \tan x = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = 2 \cot 2x$.
Therefore,$\frac{\cot x - \tan x}{\cot 2x} = \frac{2 \cot 2x}{\cot 2x} = 2$.

(B) We are given the expression: $\cos 15^{\circ} \cos 7.5^{\circ} \sin 7.5^{\circ}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we can write $\sin 7.5^{\circ} \cos 7.5^{\circ} = \frac{1}{2} \sin(2 \times 7.5^{\circ}) = \frac{1}{2} \sin 15^{\circ}$.
Substituting this into the original expression:
$\cos 15^{\circ} \times (\frac{1}{2} \sin 15^{\circ}) = \frac{1}{2} \sin 15^{\circ} \cos 15^{\circ}$.
Again,using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2} \sin(2 \times 15^{\circ}) = \frac{1}{2} \sin 30^{\circ}$.
Thus,the expression becomes $\frac{1}{2} \times (\frac{1}{2} \sin 30^{\circ}) = \frac{1}{4} \sin 30^{\circ}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,the final value is $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

(A) Given expression: $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4x}}}}$
Using the identity $1 + \cos 2\theta = 2 \cos^2 \theta$,we have $2 + 2 \cos 4x = 2(1 + \cos 4x) = 2(2 \cos^2 2x) = 4 \cos^2 2x$.
Substituting this into the innermost radical: $\sqrt{2+2 \cos 4x} = \sqrt{4 \cos^2 2x} = 2 \cos 2x$.
Now the expression becomes: $\frac{2}{\sqrt{2+\sqrt{2+2 \cos 2x}}}$.
Again,using $2 + 2 \cos 2x = 2(1 + \cos 2x) = 2(2 \cos^2 x) = 4 \cos^2 x$.
Substituting this: $\sqrt{2+2 \cos 2x} = \sqrt{4 \cos^2 x} = 2 \cos x$.
Now the expression becomes: $\frac{2}{\sqrt{2+2 \cos x}}$.
Using $2 + 2 \cos x = 2(1 + \cos x) = 2(2 \cos^2 \frac{x}{2}) = 4 \cos^2 \frac{x}{2}$.
Thus,$\sqrt{2+2 \cos x} = 2 \cos \frac{x}{2}$.
Finally,the expression simplifies to $\frac{2}{2 \cos \frac{x}{2}} = \frac{1}{\cos \frac{x}{2}} = \sec \frac{x}{2}$.

(B) Given,$\cot \frac{2x}{3} + \tan \frac{x}{3} = \operatorname{cosec} \frac{kx}{3}$.
Let $\theta = \frac{x}{3}$.
Then,$\cot 2\theta + \tan \theta = \operatorname{cosec} k\theta$.
Using the identities $\cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$\frac{\cos 2\theta}{\sin 2\theta} + \frac{\sin \theta}{\cos \theta} = \operatorname{cosec} k\theta$.
$\frac{\cos 2\theta \cos \theta + \sin 2\theta \sin \theta}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$\frac{\cos(2\theta - \theta)}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
$\frac{\cos \theta}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
$\frac{1}{\sin 2\theta} = \operatorname{cosec} k\theta$.
$\operatorname{cosec} 2\theta = \operatorname{cosec} k\theta$.
Therefore,$k = 2$.

(B) Using the identities $\cos ^2 A - \sin ^2 B = \cos(A+B) \cos(A-B)$ and $\sin ^2 A - \sin ^2 B = \sin(A+B) \sin(A-B)$:
Numerator: $\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ}+12^{\circ}) \cos(48^{\circ}-12^{\circ}) = \cos 60^{\circ} \cos 36^{\circ}$.
Denominator: $\sin ^2 24^{\circ} - \sin ^2 6^{\circ} = \sin(24^{\circ}+6^{\circ}) \sin(24^{\circ}-6^{\circ}) = \sin 30^{\circ} \sin 18^{\circ}$.
Substituting the values $\cos 60^{\circ} = \frac{1}{2}$,$\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,$\sin 30^{\circ} = \frac{1}{2}$,and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:
Expression $= \frac{(1/2) \times ((\sqrt{5}+1)/4)}{(1/2) \times ((\sqrt{5}-1)/4)} = \frac{\sqrt{5}+1}{\sqrt{5}-1}$.
Rationalizing the denominator: $\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{5+1+2\sqrt{5}}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$.
Comparing with $\frac{\alpha+\beta \sqrt{5}}{2}$,we get $\alpha = 3$ and $\beta = 1$.
Therefore,$\alpha+\beta = 3+1 = 4$.

(A) We use the trigonometric identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$.
Here,$\theta = 10^\circ = \frac{\pi}{18}$.
Then,$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ = \frac{1}{4} \sin(3 \times 10^\circ) = \frac{1}{4} \sin 30^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Now,we need to find the value of $\sin(\frac{10K\pi}{3})$.
Substituting $K = \frac{1}{8}$,we get $\sin(\frac{10 \times (1/8) \times \pi}{3}) = \sin(\frac{10\pi}{24}) = \sin(\frac{5\pi}{12})$.
Since $\frac{5\pi}{12} = 75^\circ$,we have $\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$.
$= (\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \times \frac{1}{2}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.