The Derivative of the Sine, Cosine, and Tangent Functions

The Derivative of the Sine, Cosine, and Tangent Functions

In the proofs of the theorems above, we use the following well-known limits:

(1)
\begin{align} \quad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \end{align}
(2)
\begin{align} \quad \lim_{h \to 0} \frac{\sin h}{h} = 1 \end{align}

As well as many trigonometric identities.

The Derivative of the Sine Function

Theorem 1: If $f(x) = \sin x$ then $f'(x) = \cos x$.
  • Proof:
(3)
\begin{align} \quad f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ &= \lim_{h \to 0} \left [ \frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right ] \\ &= \cos x \quad \blacksquare \end{align}

The Derivative of the Cosine Function

Theorem 2: If $f(x) = \cos x$ then $f'(x) = -\sin x$.
  • Proof:
(4)
\begin{align} \quad f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h} \\ &= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \\ &= \lim_{h \to 0} \left [ \frac{\cos x (\cos h - 1)}{h} - \frac{\sin x \sin h}{h} \right ] \\ &= -\sin x \quad \blacksquare \end{align}

The Derivative of the Tangent Function

Theorem 3: If $f(x) = \tan x$ then $f'(x) = sec^2 x$.
  • Proof: By definition, $\displaystyle{\tan x = \frac{\sin x}{\cos x}}$. Using Theorems 1 and 2 and the quotient rule we get:
(5)
\begin{align} \quad f'(x) &= \frac{[\sin x]'[\cos x] - [\cos x]'[\sin x]}{[\cos x]^2} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} \\ &= \sec^2 x \quad \blacksquare \end{align}
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