Trigonometric Functions
Recall the results of the previous post.
\[\lim_{x \rightarrow 0} \ \frac{\sin(x)}{x} = 1 \qquad\qquad \lim_{x \rightarrow 0} \ \frac{\cos(x)-1}{x} = 0\]Sine
\[\begin{align} \frac{d}{dx} \sin (x) &= \lim_{h \rightarrow 0} \ \frac{\sin(x+h) - \sin(x)}{h} \\[10pt] &= \lim_{h \rightarrow 0} \ \frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)}{h} \\[10pt] &= \lim_{h \rightarrow 0} \ \frac{\sin(x)(\cos(h)-1) + \sin(h)\cos(x)}{h} \\[10pt] &= \sin(x) \left ( \lim_{h \rightarrow 0} \ \frac{\cos(h)-1}{h} \right ) + \cos(x) \left ( \lim_{h \rightarrow 0} \ \frac{\sin(h)}{h} \right ) \\[10pt] &= \sin(x) \cdot 0 + \cos(x) \cdot 1 \\[10pt] &= \cos(x) \end{align}\]Cosine
\[\begin{align} \frac{d}{dx} \cos (x) &= \lim_{h \rightarrow 0} \ \frac{\cos(x+h) - \cos(x)}{h} \\[10pt] &= \lim_{h \rightarrow 0} \ \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} \\[10pt] &= \lim_{h \rightarrow 0} \ \frac{\cos(x)(\cos(h)-1) - \sin(x)\sin(h)}{h} \\[10pt] &= \cos(x) \left ( \lim_{h \rightarrow 0} \ \frac{\cos(h)-1}{h} \right ) - \sin(x) \left ( \lim_{h \rightarrow 0} \ \frac{\sin(h)}{h} \right ) \\[10pt] &= \cos(x) \cdot 0 - \sin(x) \cdot 1 \\[10pt] &= - \sin(x) \end{align}\]Tangent
Using the quotient rule.
\[\begin{align} \frac{d}{dx} \tan (x) &= \frac{d}{dx} \frac{\sin(x)}{\cos(x)} \\[10pt] &= \frac{\frac{d}{dx} \left [ \sin(x) \right ] \cdot \cos(x) - \sin(x) \cdot \frac{d}{dx} \left [ \cos(x) \right ]}{\left ( \cos(x) \right )^2} \\[10pt] &= \frac{\cos(x) \cdot \cos(x) - \sin(x) \cdot (-\sin(x))}{\cos^2(x)} \\[10pt] &= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} \\[10pt] &= \frac{1}{\cos^2(x)} \\[10pt] &= \sec^2(x) \\[10pt] \end{align}\]Secant
Using reciprocal rule.
\[\begin{align} \frac{d}{dx} \sec (x) &= \frac{d}{dx} \frac{1}{\cos(x)} \\[10pt] &= - \frac{\frac{d}{dx} \cos (x)}{(\cos(x))^2} \\[10pt] &= - \frac{- \sin (x)}{\cos^2(x)} \\[10pt] &= \tan(x)\sec(x) \end{align}\]Cosecant
Using reciprocal rule.
\[\begin{align} \frac{d}{dx} \csc (x) &= \frac{d}{dx} \frac{1}{\sin(x)} \\[10pt] &= - \frac{\frac{d}{dx} \sin (x)}{(\sin(x))^2} \\[10pt] &= - \frac{\cos (x)}{\sin^2(x)} \\[10pt] &= - \cot(x)\csc(x) \end{align}\]Cotangent
Using the quotient rule.
\[\begin{align} \frac{d}{dx} \cot (x) &= \frac{d}{dx} \frac{\cos(x)}{\sin(x)} \\[10pt] &= \frac{\frac{d}{dx} \left [ \cos(x) \right ] \cdot \sin(x) - \cos(x) \cdot \frac{d}{dx} \left [ \sin(x) \right ]}{\left ( \sin(x) \right )^2} \\[10pt] &= \frac{(- \sin(x)) \cdot \sin(x) - \cos(x) \cdot \cos(x)}{\sin^2(x)} \\[10pt] &= - \frac{\sin^2(x) + \cos^2(x)}{\sin^2(x)} \\[10pt] &= -\frac{1}{\sin^2(x)} \\[10pt] &= -\csc^2(x) \\[10pt] \end{align}\]