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Question

For non-negative integers n, let
f(n)=nk=0sin(k+1n+2π)sin(k+2n+2π)nk=0sin2(k+1n+2π)
Assuming cos1x takes values in [0,π], which of the following options is/are correct?
  1. f(4)=32
  2. limnf(n)=12
  3. If α=tan(cos1f(6)),  then α2+2α1=0 
  4. sin(7cos1f(5))=0


Solution

The correct options are
A f(4)=32
C If α=tan(cos1f(6)),  then α2+2α1=0 
D sin(7cos1f(5))=0
f(n)=nk=0[cos(πn+2)cos(2k+3n+2π)]nk=01cos2(k+1n+2π)

f(n)=(n+1)cos(πn+2)nk=0cos(2k+3n+2π)(n+1)nk=0cos2(k+1n+2π)

f(n)=(n+1)cos(πn+2)[cos3πn+2+cos5πn+2+...+cos2n+3n+2π](n+1)[cos2πn+2+cos4πn+2+...+cos2n+1n+2π]

f(n)=(n+1)cos(πn+2)⎢ ⎢ ⎢ ⎢sin(n+1)πn+2sinπn+2⎥ ⎥ ⎥ ⎥cos(n+3)π(n+2)(n+1)⎢ ⎢ ⎢ ⎢sin(n+1)πn+2sinπn+2⎥ ⎥ ⎥ ⎥cos2(n+2)π2(n+2)

f(n)=(n+1)cos(πn+2)⎢ ⎢ ⎢ ⎢sin(ππn+2)sinπn+2⎥ ⎥ ⎥ ⎥cos(π+π(n+2))(n+1)⎢ ⎢ ⎢ ⎢sin(ππn+2)sinπn+2⎥ ⎥ ⎥ ⎥cosπ

f(n)=(n+1)cos(πn+2)+cosπn+2(n+1)cosπ
f(n)=(n+2)cos(πn+2)(n+2)=cos(πn+2)

Option 1:
f(4)=cosπ6=32

Option 2:
limnf(n)=limncosπn+2=cos(0)12

Option 3: 
If α=tan(cos1cosπ8)=tanπ8=21
then α2+2α1=(21)2+2(21)1=0 

Option 4:
sin(7cos1cosπ7)=sinπ=0
 

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