If cos-i sin-cos 2 -i sin 2 - -cos n -i sin n -1- where - km - n 2-1 for m - Z- then the value of k is

We know that cosθ + i sinθ = e^i θ Now, (cosθ + i sinθ)(cos 2θ + i sin 2θ) ⋯ (cos nθ + i sin nθ) =1 ⇒ e^iθ× nbsp;e^i2θ⋯ nbsp; × nbsp;e^i nθ = 1 ⇒ e^i(1+2+ ...

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Byju's Answer

Standard XII

Mathematics

Euler's Representation

If cosθ+i sin...

Question

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 $k$ is

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Solution

We know that
$cos θ + i sin θ = e^{i θ}$
Now,
$(cos θ + i sin θ) (cos 2 θ + i sin 2 θ) \dots (cos n θ + i sin n θ) = 1 \Rightarrow e^{i θ} \times e^{i 2 θ} \dots \times e^{i n θ} = 1 \Rightarrow e^{i (1 + 2 + \dots + n) θ} = 1 \Rightarrow e^{i \frac{n (n + 1) θ}{2}} = 1 \Rightarrow cos (\frac{n (n + 1) θ}{2}) + i sin (\frac{n (n + 1) θ}{2}) = 1 + i 0$
Comparing real and imaginary parts, we get
$cos (\frac{n (n + 1) θ}{2}) = 1, sin (\frac{n (n + 1) θ}{2}) = 0 \Rightarrow \frac{n (n + 1) θ}{2} = 2 m π \Rightarrow θ = \frac{4 m π}{n (n + 1)} ∴ k = 4$

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If (cosθ+isinθ)(cos2θ+isin2θ)⋯(cosnθ+isinnθ)=1, where θ=kmπn2+1 for m∈Z, then the value of k is

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 $k$ is