If 1-cos-sin-1-cos-sisin-4-cosn-sin n - then n isequal to

The correct option is C 4 D' =i(1+cosθ)+sinθ = 2icos^2θ/2+2sinθ/2cosθ/2 L.H.S = [cos(ϕ/2) +isin(ϕ/2)/icos(ϕ/2) +sin(ϕ/2)]^4 = 1/i^4(cosθ + isinθ)^4 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Euler's Representation

If 1+cosθ+sin...

Question

If ${(\frac{1 + c o s θ + i s i n θ}{1 + c o s θ - i s i n θ})}^{4}$ = cosn $θ$ +isinn $θ$ , then n is

equal to

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

Open in App

Solution

The correct option is D
4

D' =i(1+cos $θ$ )+sin $θ$ = 2i $c o s^{2}$ $\frac{θ}{2}$ +2sin $\frac{θ}{2}$ cos $\frac{θ}{2}$

L.H.S = ${[\frac{c o s (\frac{ϕ}{2}) + i s i n (\frac{ϕ}{2})}{i c o s (\frac{ϕ}{2}) + s i n (\frac{ϕ}{2})}]}^{4}$

= $\frac{1}{i^{4}}$ $(c o s θ + i s i n θ)^{4}$ = cos4 $θ$ +isin4 $θ$ .

Suggest Corrections

Similar questions

Q. If

sin θ + cos θ = 1

then

(sin θ - cos θ)

is equal to

Q. The value of

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

Q. Prove by PMI:

c o s θ + c o s 2 θ + . . . . . + c o s n θ = s i n \frac{n θ}{2} . c o s θ c \frac{θ}{2} . c o s \frac{(n + 1) θ}{2}

If $m = (c o s θ - s i n θ)$ and $n = (c o s θ + s i n θ)$ then show that

$\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{2}{\sqrt{1 - t a n^{2} θ}}$

The expression $E = \frac{1 + cos θ + sin θ}{1 + cos θ - sin θ}$ can be simplified to

Join BYJU'S Learning Program

If (1+cosθ+isinθ1+cosθ−isinθ)4 = cosnθ+isinnθ , then n is equal to

The correct option is D 4 D' =i(1+cosθ)+sinθ = 2icos2θ2+2sinθ2cosθ2 L.H.S = [cos(ϕ2)+isin(ϕ2)icos(ϕ2)+sin(ϕ2)]4 = 1i4(cosθ+isinθ)4 = cos4θ+isin4θ.

If ${(\frac{1 + c o s θ + i s i n θ}{1 + c o s θ - i s i n θ})}^{4}$ = cosn $θ$ +isinn $θ$ , then n is

equal to