Sin 2π/7+sin4π/7+sin8π/7 . it`s numeric value
Arham , 10 Years ago
Grade 11
Trigonometry> Sin 2π/7+sin4π/7+sin8π/7 . it`s numeric v...
3 Answers
Nishant pandey
We seek S = sin (2π/7) + sin (4π/7) - sin (π/7)
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
Approved Last Activity: 10 Years ago
Raghu Vamshi Hemadri
We seek S = sin (2π/7) + sin (4π/7) - sin (π/7)
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
SHAIK HAFEEZUL KAREEM
We seek S = sin (2π/7) + sin (4π/7) - sin (π/7)
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.
From cos(α + β) and sin(α + β) formulas come
cos(2π/7) = c² - s² = 2c² - 1
sin(2π/7) = 2cs
sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and
sin(4π/7) = 4cs(2c²-1)
Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or
(1): 8c³ = 4c² + 4c - 1
This means any polynomial in c can be reduced to a quadratic or less.
S² = (sin(2π/7) + sin(4π/7) - sin(π/7))²
= (sin(2π/7) + sin(3π/7) - sin(π/7))²
= (2sc + s(4c²-1) - s)²
= s² * (4c² + 2c -2)²
= (1-c²) * (4c² + 2c -2)² which was the reason for squaring S: a polynomial in c,
= (4c² + 2c - 2)² - (4c³ + 2c² - 2c)² which we now reduce. First, apply (1) to 4c³ to get
= (4c² + 2c - 2)² - (4c² - ½)² which, by difference of squares is
= (8c² + 2c - 5/2) * (2c - 3/2)
= 16c³ - 8c² - 8c + 15/4 and now apply (1) a final time
= -2 + 15/4
= 7/4
Hence, the value of the original expression is S = ½ * √7
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