wiz-icon
MyQuestionIcon
MyQuestionIcon
8
You visited us 8 times! Enjoying our articles? Unlock Full Access!
Question

2n1sinπnsin2πn...sinn1nπ equals

A
n
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B n
From (xα)(x¯¯¯¯α)=x22xcosπn+1
and x2n1x21=x2n2+x2n+3+...+x+1
We get
(x22xcosπn+1)(x22xcos2πn+1)...(x22xcosn1nπ+1)=1+x+...+x2n2
Divide both sides by xn1 to obtain
(x+1x2cosπn)(x+1x2cos2πn)...(x+1x2cosn1nπ)=xnxnxx1
Now put x=cosθ+isinθ
we get
2n1(cosθcosπn)(cosθcos2πn)...(cosθcosn1nπ)=sinnθsinθ
Take limθ0
n=2n1(2sin2π2n)(2sin22π2n)...(2sin2n1nπ)
=2n1(2sinπ2nsinn12nπ)(2sin2π2nsinn22nπ)...(2sinn12nπsinπn)
Now, use
2sinkπ2nsin(nk2nπ)=sin(kπn)
we get
2n1sinπnsin2πn...sinn1nπ=n

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration by Substitution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon