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anemone
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Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.
anemone said:Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.
kaliprasad said:Note: I have used $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ and I can prove it if required
anemone said:Prove that $\cos \dfrac{\pi}{7}$ is a root of the equation $8x^3-4x^2-4x+1=0$.
kaliprasad said:Let us put $x = \cos \frac{\pi}{7}$ and see if expression is zero
we have $8x^3-4x^2-4x+1=8\cos^3 \frac{\pi}{7} - 4 \cos^2 \frac{\pi}{7} + 4 \cos \frac{\pi}{7} + 1$
$= 2(4\cos^3 \frac{\pi}{7}) - 2 (2 \cos^2 \frac{\pi}{7}) - 4 \cos \frac{\pi}{7} + 1$
$= 2 (3 cos \frac{\pi}{7} + cos \frac{3\pi}{7}) - 2 ( cos \frac{2\pi}{7} + 1) + 4 \cos \frac{\pi}{7} + 1$ using $4cos^3 x = 3 \cos\, x + cos 3x$ and $2\cos^2x = cos 2x + 1$
$= 2(cos \frac{3\pi}{7} - cos \frac{2\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{3\pi}{7} + cos \frac{5\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(cos \frac{5\pi}{7} + cos \frac{3\pi}{7} + cos \frac{\pi}{7} + \frac{1}{2})$
$= 2(-\frac{1}{2} + \frac{1}{2})= 0$
Note: I have used $(cos \frac{5\pi}{7}) + cos \frac{3\pi}{7} + cos \frac{\pi}{7} = - \frac{1}{2}$ and I can prove it if required
For a number to be a root of an equation, it means that when that number is substituted into the equation, the resulting expression will equal zero. So, if we have a cubic equation and cos(π/7) is one of its roots, it means that when we plug in cos(π/7) into the equation, the equation will equal zero.
We can prove that cos(π/7) is a root to a cubic equation by using the trigonometric identity cos(3x) = 4cos^3(x) - 3cos(x). By substituting x = π/7 into this identity, we can rearrange the equation to get cos(π/7) as a root of a cubic equation.
Yes, cos(π/7) can be a root to other types of equations, such as trigonometric equations or transcendental equations. It all depends on the equation and the values of its coefficients.
Cos(π/7) is a significant root to a cubic equation because it is one of the few values that can be expressed algebraically in terms of radicals. This means that we can write cos(π/7) as a combination of square roots, which makes it easier to work with in mathematical calculations.
Knowing that cos(π/7) is a root to a cubic equation can be useful in solving certain problems in mathematics and physics, such as finding the roots of a polynomial or solving trigonometric equations. It is also a fundamental concept in understanding the properties and behavior of trigonometric functions.