Find the exact value of $ cos(frac{pi}{9}) $ using the unit circle and trigonometric identities

To find the exact value of $ \cos(\frac{\pi}{9}) $, we can use the triple-angle identity for cosine:

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$ \cos(3\theta) = 4\cos^3(\theta) – 3\cos(\theta) $

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Letting $ \theta = \frac{\pi}{9} $, we get:

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$ \cos(\frac{\pi}{3}) = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $

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Since $ \cos(\frac{\pi}{3}) = \frac{1}{2} $, substituting in we have:

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$ \frac{1}{2} = 4\cos^3(\frac{\pi}{9}) – 3\cos(\frac{\pi}{9}) $

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Let $ x = \cos(\frac{\pi}{9}) $, then the equation becomes:

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$ \frac{1}{2} = 4x^3 – 3x $

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Multiplying through by 2 to clear the fraction:

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$ 1 = 8x^3 – 6x $

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This is a cubic equation that can be solved for $ x = \cos(\frac{\pi}{9}) $ using numerical methods or by recognizing that:

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$ \cos(\frac{\pi}{9}) = \frac{\sqrt{6} + \sqrt{2}}{4} $

To find the exact value of $ cos(frac{pi}{9}) $, we can use the triple-angle identity for cosine:

$ cos(3 heta) = 4cos^3( heta) – 3cos( heta) $

Letting $ heta = frac{pi}{9} $, we get:

$ cos(frac{pi}{3}) = 4cos^3(frac{pi}{9}) – 3cos(frac{pi}{9}) $

Since $ cos(frac{pi}{3}) = frac{1}{2} $, substituting in we have:

$ frac{1}{2} = 4cos^3(frac{pi}{9}) – 3cos(frac{pi}{9}) $

Let $ x = cos(frac{pi}{9}) $, then the equation becomes:

$ frac{1}{2} = 4x^3 – 3x $

This is a cubic equation that can be solved for $ x = cos(frac{pi}{9}) $ using numerical methods or by recognizing the solution:

$ cos(frac{pi}{9}) = frac{sqrt{6} + sqrt{2}}{4} $

The exact value of $ cos(frac{pi}{9}) $ can be found using the triple-angle identity for cosine:

$ cos(3 heta) = 4cos^3( heta) – 3cos( heta) $

Let $ heta = frac{pi}{9} $.

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