Noelanijd

Noelanijd

Answered

2022-07-23

f ( 2 π 7 ) + f ( 4 π 7 ) + f ( 6 π 7 ) = 1
let
f ( x ) = 1 1 + 2 cos x
prove that :
f ( 2 π 7 ) + f ( 4 π 7 ) + f ( 6 π 7 ) = 1
My Try :
f ( 2 π 7 ) = 1 1 + 2 cos ( 2 π 7 )
f ( 2 π 7 ) = 1 1 + 2 cos ( 4 π 7 )
f ( 2 π 7 ) = 1 1 + 2 cos ( 6 π 7 )
L = 1 1 + 2 cos ( 6 π 7 ) + 1 1 + 2 cos ( 4 π 7 ) + 1 1 + 2 cos ( 2 π 7 )
what now ?

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Answer & Explanation

Clarissa Adkins

Clarissa Adkins

Expert

2022-07-24Added 16 answers

Use factor z 7 1 into linear and quadratic factors and prove that cos ( π / 7 ) cos ( 2 π / 7 ) cos ( 3 π / 7 ) = 1 / 8
or if 7 x = 2 m π where m is any integer
sin 4 x = sin ( 2 m π 3 x ) = sin 3 x
4 sin x cos x cos 2 x = sin x ( 4 sin 2 x 3 )
4 sin x cos x ( 2 cos 2 x 1 ) = sin x { 4 ( 1 cos 2 x ) 1 }
So, the roots of 4 cos x ( 2 cos 2 x 1 ) = 4 ( 1 cos 2 x ) 1 8 cos 3 x + 4 cos 2 x 4 cos x 3 = 0         ( 1 )
are 7 x = 2 m π where m ± 1 , ± 2 , ± 3 ( mod 7 )
Now if u = 1 1 + 2 cos x cos x = ?
Replace the values of cos x in ( 1 )

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Levi Rasmussen

Levi Rasmussen

Expert

2022-07-25Added 6 answers

Let 2 cos 2 π 7 = a, 2 cos 4 π 7 = b and 2 cos 6 π 7 = c
Hence,
a + b + c = 2 sin π 7 cos 2 π 7 + 2 sin π 7 cos 4 π 7 + 2 sin π 7 cos 6 π 7 sin π 7 =
= sin 3 π 7 sin π 7 + sin 5 π 7 sin 3 π 7 + sin 7 π 7 sin 5 π 7 sin π 7 = 1 ;
a b + a c + b c = 4 ( cos 2 π 7 cos 4 π 7 + cos 2 π 7 cos 6 π 7 + cos 4 π 7 cos 6 π 7 ) =
= 2 ( cos 6 π 7 + cos 2 π 7 + cos 6 π 7 + cos 4 π 7 + cos 4 π 7 + cos 2 π 7 ) = 2
and
a b c = 8 cos 2 π 7 cos 4 π 7 cos 6 π 7 = 8 sin 2 π 7 cos 2 π 7 cos 4 π 7 cos 8 π 7 sin 2 π 7 = sin 16 π 7 sin 2 π 7 = 1.
Id est,
f ( 2 π 7 ) + f ( 4 π 7 ) + f ( 6 π 7 ) = c y c 1 1 + a = c y c ( a + 1 ) ( b + 1 ) c y c ( 1 + a ) =
= a b + a c + b c + 2 ( a + b + c ) + 3 1 + a + b + c + a b + a c + b c + a b c = 2 2 + 3 1 1 2 + 1 = 1.
Done!

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