Latest revision as of 17:45, 20 February 2018

Problem 1

Solution

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Problem 2

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Problem 3

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Trigonometric Identities

• $ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $
• $ \cos\left(x\right) + \cos\left(x - \frac{2\pi}{3}\right) + \cos\left(x + \frac{2\pi}{3}\right) = 0 $
• $ \sin\left(x\right) + \sin\left(x - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) = 0 $
• $ \cos\left(x\right) \cos\left(y\right) + \cos\left(x - \frac{2\pi}{3}\right) \cos\left(y - \frac{2\pi}{3}\right) + \cos\left(x + \frac{2\pi}{3}\right) \cos\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \cos(x - y) $
• $ \sin\left(x\right) \sin\left(y\right) + \sin\left(x - \frac{2\pi}{3}\right) \sin\left(y - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) \sin\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \cos(x - y) $
• $ \sin\left(x\right) \cos\left(y\right) + \sin\left(x - \frac{2\pi}{3}\right) \cos\left(y - \frac{2\pi}{3}\right) + \sin\left(x + \frac{2\pi}{3}\right) \cos\left(y + \frac{2\pi}{3}\right) = \frac{3}{2} \sin(x - y) $

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