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Trigonometric Ratios for Sum of Two Angles

If fnθ =cosθ2...

Question

If $f_{n} (θ) = \frac{cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2}}{sin \frac{θ}{2} + sin 2 θ + sin \frac{7 θ}{2} + \dots + sin \frac{(3 n - 2) θ}{2}}$ , then match the entries of $Column-I$ with their corresponding values in $Column-II .$

$\begin{matrix} Column-I & Column-II (A) & f_{1} (\frac{π}{2}) & (P) & 0 (B) & f_{3} (\frac{3 π}{16}) & (Q) & \sqrt{2} - 1 (C) & f_{4} (\frac{2 π}{11}) & (R) & \sqrt{2} + 1 (D) & f_{5} (\frac{π}{28}) & (S) & 1 (T) & 2 - \sqrt{3} \end{matrix}$

(A) \to (P) (B) \to (Q) (C) \to (R) (D) \to (S)

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(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)

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(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (T)

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(A) \to (T) (B) \to (R) (C) \to (P) (D) \to (R)

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Solution

The correct option is B $(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)$

$cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2} = cos \frac{θ}{2} + cos (\frac{θ}{2} + \frac{3 θ}{2}) + cos (\frac{θ}{2} + 2 (\frac{3 θ}{2})) + \dots + cos (\frac{θ}{2} + (n - 1) (\frac{3 θ}{2}))$

Here, $A = \frac{θ}{2}, D = \frac{3 θ}{2}$
Number of trem $= n$
So,
$cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2} = \frac{sin n (\frac{3 θ}{4})}{sin \frac{3 θ}{4}} ⎡ ⎢ ⎢ ⎢ ⎣ cos ⎛ ⎜ ⎜ ⎜ ⎝ \frac{θ + (n - 1) \frac{3 θ}{2}}{2} ⎞ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦ = \frac{sin (\frac{3 n θ}{4})}{sin \frac{3 θ}{4}} [cos \frac{(3 n - 1) θ}{4}]$

Similarly,
$sin \frac{θ}{2} + sin 2 θ + sin \frac{7 θ}{2} + \dots + sin \frac{(3 n - 2) θ}{2} = \frac{sin n (\frac{3 θ}{4})}{sin \frac{3 θ}{4}} ⎡ ⎢ ⎢ ⎢ ⎣ sin ⎛ ⎜ ⎜ ⎜ ⎝ \frac{θ + (n - 1) \frac{3 θ}{2}}{2} ⎞ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦ = \frac{sin (\frac{3 n θ}{4})}{sin \frac{3 θ}{4}} [sin \frac{(3 n - 1) θ}{4}]$

Therefore,
$f_{n} (θ) = cot \frac{(3 n - 1) θ}{4}$

Now,
$f_{1} (\frac{π}{2}) = cot \frac{2 π}{4 \cdot 2} = 1 f_{3} (\frac{3 π}{16}) = cot \frac{8 \cdot 3 π}{4 \cdot 16} = cot \frac{3 π}{8} = \sqrt{2} - 1 f_{4} (\frac{2 π}{11}) = cot \frac{11 \cdot 2 π}{4 \cdot 11} = 0 f_{5} (\frac{π}{28}) = cot \frac{14 π}{4 \cdot 28} = cot \frac{π}{8} = \sqrt{2} + 1$

Hence,
$(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)$

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Similar questions

f_{n} (θ) = \frac{cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2}}{sin \frac{θ}{2} + sin 2 θ + sin \frac{7 θ}{2} + \dots + sin \frac{(3 n - 2) θ}{2}}

Column-I

Column-II .

\begin{matrix} Column-I & Column-II (A) & f_{1} (\frac{π}{2}) & (P) & 0 (B) & f_{3} (\frac{3 π}{16}) & (Q) & \sqrt{2} - 1 (C) & f_{4} (\frac{2 π}{11}) & (R) & \sqrt{2} + 1 (D) & f_{5} (\frac{π}{28}) & (S) & 1 (T) & 2 - \sqrt{3} \end{matrix}

f_{n} (θ) = \frac{cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2}}{sin \frac{θ}{2} + sin 2 θ + sin \frac{7 θ}{2} + \dots + sin \frac{(3 n - 2) θ}{2}}

Column-I

Column-II .

\begin{matrix} Column-I & Column-II (A) & f_{1} (\frac{π}{2}) & (P) & 0 (B) & f_{3} (\frac{3 π}{16}) & (Q) & \sqrt{2} - 1 (C) & f_{4} (\frac{2 π}{11}) & (R) & \sqrt{2} + 1 (D) & f_{5} (\frac{π}{28}) & (S) & 1 (T) & 2 - \sqrt{3} \end{matrix}

f_{n} (θ) = \frac{cos \frac{θ}{2} + cos 2 θ + cos \frac{7 θ}{2} + \dots + cos \frac{(3 n - 2) θ}{2}}{sin \frac{θ}{2} + sin 2 θ + sin \frac{7 θ}{2} + \dots + sin \frac{(3 n - 2) θ}{2}},

x^{2} + a | x | + 1 = 0

a

\begin{matrix} List I & List II (A) & No real roots & (P) & a < - 2 (B) & Two real roots & (Q) & ϕ (C) & Three real roots & (R) & a = - 2 (D) & Four distinct real roots & (S) & a \geq 0 (T) & a = 2 \end{matrix}

\begin{matrix} Column I & Column II a. sin (410^{\circ} - A) cos (400^{\circ} + A) + cos (410^{\circ} - A) sin (400^{\circ} + A) & p. -1 b. \frac{{cos}^{2} 1^{\circ} - {cos}^{2} 2^{\circ}}{2 sin 3^{\circ} sin 1^{\circ}} is equal to & q. 1 c. sin (- 870^{\circ}) + cosec (- 660^{\circ}) + tan (- 855^{\circ}) + 2 cot (840^{\circ}) + cos (480^{\circ}) + sec (900^{\circ}) & r. \frac{1}{2} \end{matrix}

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