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Differentiation of Inverse Trigonometric Functions

If cos 5 θ ...

Question

If $cos 5 θ = a_{0} {cos}^{5} θ + a_{1} {cos}^{4} θ + a_{2} {cos}^{3} θ + a_{3} {cos}^{2} θ + a_{4} cos θ + a_{5}$ then the value of $(a_{1} + a_{3} + a_{5})$ is equal to ?

A

- 1

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B

0

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C

12

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D

- 16

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Solution

The correct option is B $0$
$cos 5 θ = a_{0} {cos}^{5} θ + a_{1} {cos}^{4} θ + a_{2} {cos}^{3} θ + a_{3} {cos}^{2} θ + a_{4} cos θ + a_{5}$
Substituing $θ = 0$ :
$1 = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$ (1)
Substituing $θ = π$ :
$- 1 = - a_{0} + a_{1} - a_{2} + a_{3} - a_{4} + a_{5}$ (2)
Adding (1) and (2):
$0 = 2 (a_{1} + a_{3} + a_{5}) ⟹ a_{1} + a_{3} + a_{5} = 0$

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Similar questions

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

\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}) =

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

A . P .

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}, . . . ., a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

For an increasing A.P. $a_{1}, a_{2}, a_{3} . . . a_{n}$ ,
if $a_{1} + a_{3} + a_{5} = - 12$ and $a_{1} \cdot a_{3} \cdot a_{5} = 80$ , then which of the following is true.

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