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Determinant

If An=cosnθ...

Question

If $A_{n} = {cos}^{n} θ + {sin}^{n} θ, n \in N$ and $θ \in R$

Then, the value of $⎛ ⎜ ⎜ ⎜ ⎝ \frac{A_{6} + A_{4} - \frac{3}{4}}{A_{6} - A_{4} + \frac{1}{4}} ⎞ ⎟ ⎟ ⎟ ⎠$ is equal to

A

15

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B

4

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C

5

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D

3

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Solution

The correct option is A $5$
$A_{n} = {cos}^{n} θ + {sin}^{n} θ$
$A_{6} = {cos}^{6} θ + {sin}^{6} θ$
$= 1 - 3 {sin}^{2} θ {cos}^{2} θ$
$= 1 - \frac{3}{4} {sin}^{2} 2 θ$
$A_{4}$ $= {cos}^{4} θ + {sin}^{4} θ$
$= 1 - 2 {sin}^{2} θ {cos}^{2} θ$
$= 1 - \frac{1}{2} {sin}^{2} 2 θ$
$(A_{6} + A_{4})$ $= 1 - \frac{3}{4} {sin}^{2} 2 θ + 1 - \frac{1}{2} {sin}^{2} 2 θ$
$= 2 - \frac{5}{4} {sin}^{2} 2 θ$
$(A_{6} - A_{4}) = 1 - \frac{3}{4} {sin}^{2} 2 θ - 1 + \frac{1}{2} {sin}^{2} 2 θ$
$= - \frac{1}{4} {sin}^{2} 2 θ$
$⎛ ⎜ ⎜ ⎜ ⎝ \frac{A_{6} + A_{4} - \frac{3}{4}}{A_{6} - A_{4} + \frac{1}{4}} ⎞ ⎟ ⎟ ⎟ ⎠$ $= \frac{2 - \frac{5}{4} {sin}^{2} 2 θ - \frac{3}{4}}{- \frac{1}{4} {sin}^{2} 2 θ + \frac{1}{4}}$
$= \frac{\frac{5}{4} (1 - {sin}^{2} 2 θ)}{\frac{1}{4} (1 - {sin}^{2} 2 θ)}$
$= 5$

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Similar questions

A_{n} = {cos}^{n} θ + {sin}^{n} θ, n \in N

θ \in R

A_{n - 4} - A_{n - 2} = {sin}^{2} θ {cos}^{2} θ A_{λ}

λ

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

0 \leq a < j

j = 2, 4, 5, 6, 7.

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

0 \leq a_{j} < j

j = 2, 3, 4, 5, 6, 7

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

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