Evaluate sin(2pi/5): Solution Attempted

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In summary, the homework statement is that sin(2pi/5) = ((sqrt(10) + (2sqrt(5)))/4) + (1 - ((sqrt(10) + (2sqrt(5)))/4)) which can be solved by first solving for cos(2pi/5) and then using a quadratic equation to find x.
  • #1
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Homework Statement


Show that sin(2pi/5) = ((sqrt(10) + (2sqrt(5)))/4)


Homework Equations





The Attempt at a Solution


i showed that cos(2pi/5) + isin(2pi/5) was a primitive 5th root of unity, and that cos(2pi/5) = (-1 + sqrt(5))/4 but i cannot figure out how to do this one any ideas?
 
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  • #2
i also can use the fact that x^5 - 1 = 0 and i did a substitution to prove that cos(2pi/5) = (-1 + sqrt(5))/4 using the fact that x^5 - 1 + ( x - 1)(x^4 + x^3 + x^2 + x + 1) and x-1 is nonzero so the second equation on rhs has to be zero and then i used a variable substitution but i don't know how to go about proving the sin(2pi/5) part
 
  • #3
Well

[tex] \sin^2 \frac{2\pi}{5} + \cos^2 \frac{2\pi}{5} =1 [/tex]

, obviously.
 
  • #5
How did you get cos(2pi/5) ? I couldn't factor any polynomial.
 
  • #6
you divide x^4 + x^3 + x^2 + x + 1 by x^2 because it is non zero this is valid. so we get x^2 + x + 1 + 1/x + 1/x^2. then say y = x + 1/x and plugging this back in you get y^2 + y -1 = 0
then use the quadratic formula to get (-1 + sqrt(5))/2. back to y we know that x = cos(2pi/5) + isin(2pi/5) so add that to x^-1 and you end up getting 2cos(2pi/5) = (-1 + sqrt(5))/2
 
  • #7
Thanks for the solution. Really nice one.
 
  • #8
Hi bigubau I used a much more mundane method.

Denoting [itex]\frac{2 \pi}{5} = \theta[/itex] we have,

[tex] Re\{(\cos \theta + i \sin \theta)^5\} = 1 [/tex]

[tex] Im\{(\cos \theta + i \sin \theta)^5\} = 0 [/tex]

Expanding the second of these two equations and denoting [itex]x = \cos \theta[/itex] (and using [itex]\sin^2 \theta = 1 - x^2 [/itex] where appropriate) gives,

[tex] i \sin \theta \, \left[ 5x^4 - 10 x^2 (1-x^2) + (1-x^2)^2 \right] = 0[/tex]

which reduces to a quadratic in [itex]x^2[/itex].

[tex] 16x^4 - 12 x^2 + 1 = 0[/tex]

Solving gives,

[tex]x^2 = \frac{6 \pm 2 \sqrt{5}} {16} = \frac{(\sqrt{5} \pm 1)^2}{4^2}[/tex]

[tex]x = \frac{\sqrt{5} - 1}{4}[/tex]

BTW. Something a bit weak here, I selected the plus or minus by comparing the floating point approx of the surd with that of the cosine. I couldn't think of a better way but if someone else can then please let me know. :)
 
Last edited:

What is the value of sin(2pi/5)?

The value of sin(2pi/5) is approximately 0.5878.

How do you evaluate sin(2pi/5)?

To evaluate sin(2pi/5), you can use the unit circle or a calculator. On the unit circle, you can find the sine value by looking at the y-coordinate of the point where the angle 2pi/5 intersects the unit circle. Using a calculator, you can simply enter the angle in radians and press the sine button to get the value.

What is the exact value of sin(2pi/5)?

The exact value of sin(2pi/5) cannot be expressed in terms of integers or commonly used constants. It can be approximated as a decimal or expressed as a radical using trigonometric identities.

Why is the value of sin(2pi/5) important?

The value of sin(2pi/5) is important in mathematics because it is a special angle that can be used to derive various trigonometric identities and solve problems in geometry, physics, and other fields. It is also used in the study of periodic functions and Fourier series.

Can you explain the solution for evaluating sin(2pi/5)?

To evaluate sin(2pi/5), we can use the formula sin(2x) = 2sin(x)cos(x). Substituting x = pi/5, we get sin(2pi/5) = 2sin(pi/5)cos(pi/5). Using the half-angle identities, we can express sin(pi/5) and cos(pi/5) in terms of the golden ratio (1+√5)/2. Simplifying the expression, we get sin(2pi/5) = (1+√5)/4, which is approximately 0.5878.

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