value of sin2π/7+sin4π/7+sin8π/7=
Answers
Answer:
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
Answer:
How can I prove ~ sin 2π/7 + sin 4π/7 + sin 8π/7 = √7/2?
Let x=2π7⟹7x=2π
∴3x=2π−4x
So, cosx+cos2x+cos4x
=12sinx2(2sinx2cosx+2sinx2cos2x+2sinx2cos4x)
=12sinx2(sin3x2−sinx2+sin5x2−sin3x2+sin9x2−sin7x2)
=12sinx2(−sinx2+sin9x2+sin5x2−sin2π2)[∵7x=2π]
=12sinx2(−sinx2+2sin7x2cosx)[∵sinπ=0]
=12sinx2(−sinx2+0)[∵sin7x2=sinπ=0]
∴cosx+cos2x+cos4x=−12(i)
Now,
(sinx+sin2x+sin4x)2+(cosx+cos2x+cos4x)2
=1+1+1+2(sinxsin2x+sin2xsin4x+sin4xsinx)+2(cosxcos2x+cos2xcos4x+cos4xcosx)
=3+2[cos(2x−x)+cos(4x−2x)+cos(4x−x)]
=3+2[cosx+cos2x+cos4x]
[∵cos3x=cos(2π−4x)=cos4x]
=3+2⋅(−12)[From Eq. (i)]
=3−1
⟹(sinx+sin2x+sin4x)2+(−12)2=2
⟹(sinx+sin2x+sin4x)2=2−14=74
∴sinx+sin2x+sin4x=7–√2.†
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