The general value of the real angle - which satisfies the equation- cos-i sin-cos 2 -i sin 2 - -cos n -i sin n -1 is given by-assuming k is an integerA- 2 k - n -21 kB- 4 K - n n -1C- 4 k - n -1D- ok-nn-1

The correct option is B 4k πn(n+1)Using De Moivre's formula e^i θ· nbsp; e^i 2 θ⋯ e^i n θ = 1 ⇒ e^i θn(n+1)/2 = 1 ⇒n(n+1) θ2 = 2k π, k ∈ I ⇒θ = 4k πn(n+1), k ...

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Standard XII

Mathematics

Quadratic Identity

The general v...

Question

The general value of the real angle $θ,$ which satisfies the equation, $(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) . . . (cos n θ + i sin n θ) = 1$ is given by, (assuming $k$ is an integer)

\frac{2 k π}{n + 2}

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\frac{4 k π}{n (n + 1)}

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\frac{4 k π}{n + 1}

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\frac{6 k π}{n (n + 1)}

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Solution

The correct option is B $\frac{4 k π}{n (n + 1)}$
Using De Moivre's formula
$e^{i θ} \cdot e^{i 2 θ} \dots e^{i n θ} = 1 \Rightarrow e^{i θ \frac{n (n + 1)}{2}} = 1 \Rightarrow \frac{n (n + 1) θ}{2} = 2 k π, k \in I \Rightarrow θ = \frac{4 k π}{n (n + 1)}, k \in I$

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Similar questions

Q. The value of Z satisfying the equation

l o g Z + l o g z^{2} + l o g z^{3} + . . . . . + l o g z^{n} = 0

Q. If

(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) \dots (cos n θ + i sin n θ) = 1

, where

θ = \frac{k m π}{n^{2} + 1}

for

m \in Z

, then the value of

k

Q. If

(c o s θ + i s i n θ) (c o s 2 θ + i s i n θ)

....

(c o s n θ + i s i n n θ) = 1

, then the value of

θ

Q. Find the value of

(c o s θ + i s i n θ)^{n}

where

θ ϵ R, n ϵ I, i = \sqrt{- 1}

Q. If

{[\frac{1 + c o s θ + i s i n θ}{s i n θ + i (1 + c o s θ)}]}^{4} = c o s n θ + i s i n n θ

, then n is ______.

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The general value of the real angle θ, which satisfies the equation, (cosθ+isinθ)(cos2θ+isin2θ)...(cosnθ+isinnθ)=1 is given by, (assuming k is an integer)

The correct option is B 4kπn(n+1)Using De Moivre's formula eiθ⋅ei2θ⋯einθ=1⇒eiθn(n+1)2=1⇒n(n+1)θ2=2kπ,k∈I⇒θ=4kπn(n+1),k∈I

The general value of the real angle $θ,$ which satisfies the equation, $(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) . . . (cos n θ + i sin n θ) = 1$ is given by, (assuming $k$ is an integer)

The correct option is B $\frac{4 k π}{n (n + 1)}$
Using De Moivre's formula
$e^{i θ} \cdot e^{i 2 θ} \dots e^{i n θ} = 1 \Rightarrow e^{i θ \frac{n (n + 1)}{2}} = 1 \Rightarrow \frac{n (n + 1) θ}{2} = 2 k π, k \in I \Rightarrow θ = \frac{4 k π}{n (n + 1)}, k \in I$