Find the value of cos-isin- n where - R - n - I - i -1A- cosn-i sin n -B- cosn-i n sin-C- ncos-i sin-D- n cos-i sin n-

The correct option is A Given (cos θ + i sin θ)^n…… (1) We know e^iθ = cosθ + i sinθ We can replace in equation (1) From equation 1 we get, ( e^iθ ) ^n =e^inθ ...

You visited us 20 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

De-Moivre's Theorem

Find the valu...

Question

Find the value of $(c o s θ + i s i n θ)^{n}$ where $θ ϵ R, n ϵ I, i = \sqrt{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
Given $(c o s θ + i s i n θ)^{n} \dots \dots$ (1)
We know $e^{i θ} = c o s θ + i s i n θ$
We can replace in equation (1)
From equation 1 we get, ${(e^{i θ})}^{n}$
= $e^{i n θ}$
This can be written as $(c o s n θ + i s i n n θ)$

Suggest Corrections

Similar questions

Q. Using De Moivre's theorem prove that

{(\frac{1 + cos θ + i sin θ}{1 + cos θ - i sin θ})}^{n} = cos n θ + i sin n θ,

where

i = \sqrt{- 1}

Q. The value of

(1 + cos θ + i sin θ)^{n} + (1 + cos θ - i sin θ)^{n}

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. Prove that

{(1 + cos θ + i sin θ)}^{n} + {(1 + cos θ - i sin θ)}^{n} = 2^{n + 1} {cos}^{n} \frac{θ}{2} . cos \frac{n θ}{2}

, here

n \in N

Express the following in a + ib form:

(i)

{(\frac{cos θ + i sin θ}{sin θ + i cos θ})}^{4} = cos n θ + i sin n θ

. Find the value of

n

Join BYJU'S Learning Program

Find the value of (cosθ+isinθ)n where θϵR,nϵI,i=√−1

The correct option is C Given (cosθ+isinθ)n…… (1) We know eiθ=cosθ+isinθ We can replace in equation (1) From equation 1 we get, (eiθ)n =einθ This can be written as (cosnθ+isinnθ)

Find the value of $(c o s θ + i s i n θ)^{n}$ where $θ ϵ R, n ϵ I, i = \sqrt{- 1}$

The correct option is C
Given $(c o s θ + i s i n θ)^{n} \dots \dots$ (1)
We know $e^{i θ} = c o s θ + i s i n θ$
We can replace in equation (1)
From equation 1 we get, ${(e^{i θ})}^{n}$
= $e^{i n θ}$
This can be written as $(c o s n θ + i s i n n θ)$