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Give an alternate proof of the continuity of the sine and cosine functions

  1. Use the inequality

        \[ | \sin x | < |x|, \qquad \text{for } 0 < |x| < \frac{\pi}{2}, \]

    to prove that the sine function is continuous at 0.

  2. Recall the trig identity,

        \[ \cos 2x = 1 - 2 \sin^2 x. \]

    Use this and part (a) to prove that the cosine function is continuous at 0.

  3. Use the formulas for sine and cosine of a sum to prove that the sine and cosine functions are continuous for all x \in \mathbb{R}.

  1. Proof. To show \sin is continuous at 0, we must show that \lim_{x \to 0} \sin x = \sin 0 = 0. We show this limit is zero directly from the epsilon-delta definition of the limit, i.e., given arbitrary positive \varepsilon, we have |\sin x| < \varepsilon whenever |x| < \delta. Let \varepsilon be an arbitrary number greater than 0, \varepsilon > 0. Then, let \delta = \varepsilon. Using the given inequality we have,

        \[ | \sin x | < |x| < \delta = \varepsilon, \qquad \text{whenver } 0 < |x| < \delta. \]

    Thus, \sin x is continuous at 0. \qquad \blacksquare

  2. Proof. First, using the given trig identity we have

        \[ \cos (2x) = 1 - 2 \sin^2 x \quad \implies \quad \cos x = 1 - 2 \sin^2 \frac{x}{2}. \]

    Thus,

        \begin{align*}  \lim_{x \to 0} \cos x &= \lim_{x \to 0} \left( 1 - 2 \sin^2 \frac{x}{2} \right) \\  &= 1 - 2 \cdot \lim_{x \to 0} \left( \sin \left( \frac{x}{2} \right) \sin \left( \frac{x}{2} \right) \right) \\  &= 1 = \cos 0. \end{align*}

    Thus, cosine is continuous at 0.

  3. Proof. Finally, to show sine and cosine are continuous for all x \in \mathbb{R}, we show that \lim_{x \to h} \sin x = sin h, and \lim_{x \to h} \cos x = cos h. First, we recall the formulas for the sine and cosine of a sum,

        \begin{align*}  \sin (x+h) &= \sin x \cos h + \sin h \cos x \\  \cos (x+h) &= \cos x \cos h - \sin x \sin h. \end{align*}

    So, we compute the limits

        \begin{align*}  \lim_{x \to h} \sin x &= \lim_{x \to 0} \sin (x+h) = \lim_{x \to 0} (\sin x \cos h + \sin h \cos x) = \sin h \\  \lim_{x \to h} \cos x &= \lim_{x \to 0} \cos (x+h) = \lim_{x \to 0} (\cos x \cos h - \sin x \sin ) = \cos h. \end{align*}

    Therefore, sine and cosine are continuous for all x \in \mathbb{R}. \qquad \blacksquare

2 comments

  1. H says:

    you can prove a by the squeeze principle and the definition of continuity:

    1. sin(x) is defined at 0 (sin(0)=0).

    2. we have |sin x|<|x|
    -|x|<sin x<|x|
    lim |x| = 0 , as x approaches to 0
    lim -|x| = 0, as x approaches to 0
    thus , lim sin(x)=0 as x approaches to 0 (by the squeeze principle)

    Therefore, sin(x) is continuous at 0.

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