1-cos-1 -i sin-cos-isin-n-A- cos n -i sin n -B- cos n -i sin n -C- sin n -i cos n -D- sin n -i cos n -

The correct option is B nbsp; nbsp;[2cos^2(ϕ/2) +2isin(ϕ/2)cos(ϕ/2)/2cos^2(ϕ/2) 2isin(ϕ/2)cos(ϕ/2)]^n = nbsp;[2cos(ϕ/2) +2isin(ϕ/2)/2cos(ϕ/2) 2isin( ...

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Byju's Answer

Standard XI

Mathematics

Arithmetic Progression

1+cosϕ1 ..ϕ+i...

Question

${(\frac{1 + c o s ϕ + i s i n ϕ}{1 + c o s ϕ - i s i n ϕ})}^{n}$ =

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Solution

The correct option is B

${[\frac{2 c o s^{2} (\frac{ϕ}{2}) + 2 i s i n (\frac{ϕ}{2}) c o s (\frac{ϕ}{2})}{2 c o s^{2} (\frac{ϕ}{2}) - 2 i s i n (\frac{ϕ}{2}) c o s (\frac{ϕ}{2})}]}^{n}$

= ${[\frac{2 c o s (\frac{ϕ}{2}) + 2 i s i n (\frac{ϕ}{2})}{2 c o s (\frac{ϕ}{2}) - 2 i s i n (\frac{ϕ}{2})}]}^{n}$ = $[\frac{e^{i (\frac{ϕ}{2})}}{e^{- i (\frac{ϕ}{2})}}]$ = $(e^{i ϕ})^{n}$

=cosn $ϕ$ + isinn $ϕ$ .

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Similar questions

Q. Find the value of

r

{(\frac{1 + cos ϕ + i sin ϕ}{1 + cos ϕ - i sin ϕ})}^{n} = r (cos n ϕ + i sin n ϕ)

Q. Perform the indicated operations.

(1 + c o s ϕ + i s i n ϕ)^{n} .

a = cos θ + i sin θ, b = cos ϕ + i sin ϕ, c = cos ψ + i sin ψ

and

\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 2

, then

sin (θ - ϕ) + sin (ϕ - ψ) + sin (ψ - θ)

equals

Q. If

θ, ϕ

are acute,

sin θ = \frac{1}{2}, cos ϕ = \frac{1}{3},

then

(θ + ϕ) \in

Q. (a)

1 + i tan α (- π < α < π, α \neq \pm \frac{π}{2})

(b) Find the modulus and argument of the complex number

z_{1} = z^{2} - z

z = cos ϕ + i sin ϕ

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(1+cosϕ+isinϕ1+cosϕ−isinϕ)n =

The correct option is B [2cos2(ϕ2)+2isin(ϕ2)cos(ϕ2)2cos2(ϕ2)−2isin(ϕ2)cos(ϕ2)]n = [2cos(ϕ2)+2isin(ϕ2)2cos(ϕ2)−2isin(ϕ2)]n = [ei(ϕ2)e−i(ϕ2)]= (eiϕ)n =cosnϕ + isinnϕ.

${(\frac{1 + c o s ϕ + i s i n ϕ}{1 + c o s ϕ - i s i n ϕ})}^{n}$ =