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Trigonometric identity

Let a, b, c be positive real numbers such that a+b+c=\pi. Prove the following trigonometric identities:

  1. \displaystyle \sin a + \sin b + \sin c = 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}
  2. \displaystyle \cos a + \cos b + \cos c = 1+4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2}

Solution

  1. We have successively:

        \begin{align*} \sin a + \sin b +\sin c &= 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2} + \sin c \\ &=2 \sin \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \cos \frac{a-b}{2} + 2 \sin \frac{c}{2} \cos \frac{c}{2} \\ &=2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2} + \sin \frac{c}{2} \right ) \\ &= 2 \cos \frac{c}{2} \left ( \cos \frac{a-b}{2}+ \cos \frac{a+b}{2} \right ) \\ &= 2 \cos \frac{c}{2} \cdot 2 \cos \frac{a}{2} \cos \frac{b}{2} \\ &= 4 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2} \end{align*}

  2. Similarly, we have successively:

        \begin{align*} \cos a + \cos b + \cos c &=2 \cos \frac{a+b}{2} \cos \frac{a- b}{2} + \cos c \\ &=2 \cos \left ( \frac{\pi}{2} - \frac{c}{2} \right ) \cos \frac{a-b}{2} + \cos c \\ &=2 \sin \frac{c}{2} \cos \frac{a-b}{2} + \left ( 1 - 2 \sin^2 \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left (\cos \frac{a-b}{2} - \sin \frac{c}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \left [ \cos \frac{a-b}{2} - \sin \left ( \frac{\pi}{2} -\frac{a+b}{2} \right ) \right ] \\ &=1 + 2 \sin \frac{c}{2} \left ( \cos \frac{a-b}{2} - \cos \frac{a+b}{2} \right ) \\ &=1 + 2 \sin \frac{c}{2} \cdot 2 \sin \frac{a}{2} \sin \frac{b}{2} \\ &=1 + 4 \sin \frac{a}{2} \sin \frac{b}{2} \sin \frac{c}{2} \end{align*}

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