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Prove some particular values for sine and cosine

by RoRi

Prove the following:

  1. \sin \frac{\pi}{6} = \frac{1}{2}, \ \ \cos \frac{\pi}{6} = \frac{1}{2} \sqrt{3}.
  2. \sin \frac{\pi}{3} = \frac{1}{2} \sqrt{3}, \ \ \cos \frac{\pi}{6} = \frac{1}{2}.
  3. \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{2} \sqrt{2}.
  1. Proof. We use the triple angle identity of the sine function,

        \[ \sin \frac{\pi}{2} = \sin \left( 3 \cdot \frac{\pi}{6} \right) = 3 \sin \frac{pi}{6} - 4 \sin^3 \frac{\pi}{6}. \]

    Then, since \sin \frac{\pi}{2} = 1, we have,

        \[ 3 \sin \frac{\pi}{6} - 4 \sin^3 \frac{pi}{6} = 1 \ \ \implies \ \ \sin \frac{\pi}{6} = \frac{1}{2}. \]

    For the cosine part, we use the triple angle identity for the cosine (the second version in the exercise),

        \[ \cos \frac{\pi}{2} = \cos \left( 3 \cdot \frac{\pi}{6} \right) = 4 \cos^3 \frac{\pi}{6} - 3 \cos \frac{\pi}{6}. \]

    Thus,

        \[ 4 \cos^3 \frac{\pi}{6} - 3 \cos \frac{\pi}{6} = 0 \ \ \implies \ \ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}. \]

    In solving both of these cubics, we used that \sin \frac{pi}{6} \neq 0 and \cos \frac{\pi}{6} \neq 0 which we know since we determined all real for which sine and cosine are zero. \qquad \blacksquare

  2. Here we use part (a) and the co-relations (Thm 2.3, part d),

        \[ \cos \left( \frac{\pi}{2} + x \right) = - \sin x \implies \cos \frac{\pi}{6} = - \sin \left(-\frac{\pi}{3} \right) \implies \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}. \]

    Similarly,

        \[ \sin \left(\frac{\pi}{2} + x \right) = \cos x \implies \sin \frac{\pi}{6} = \cos \left(-\frac{\pi}{3} \right) = \cos \frac{\pi}{3} \implies \cos \frac{\pi}{3} = \frac{1}{2}. \qquad \blacksquare \]

  3. Proof. First, by the double angle identity for sine,

        \[ 1 = \sin \frac{\pi}{2} = \sin \left(2 \cdot \frac{\pi}{4} \right) = 2 \sin \frac{\pi}{4} \cos \frac{\pi}{4}. \]

    And by the double angle identity for the cosine,

        \[ 0 = \cos \frac{\pi}{2} = \cos \left( 2 \cdot \frac{\pi}{4} \right) = 1 - 2 \sin^2 \frac{\pi}{4}. \]

    Using the second equation first, we have,

        \[ 1 - 2 \sin^2 \frac{\pi}{4} = 0 \implies \sin^2 \frac{\pi}{4} = \frac{1}{2} \implies \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \]

    Then, using this value in the first equation,

        \[ 2 \sin \frac{\pi}{4} \cos \frac{\pi}{4} = 1 \implies \sqrt{2} \cos \frac{\pi}{4} = 1 \implies \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \]

    Thus, \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \qquad \blacksquare

2 comments

    • Navadeep Reddy says:

      sin (π/6) ≠ -1 .
      since in the previous problem [2-c] we proved that
      sin(x) = -1 for x = 3π/2 + 2nπ
      and (π/6) ≠ 3π/2 + 2nπ for any n∈Z.

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