Two Fundamental Trigonometric Limits:
When \(\theta\) is very small, \(\sin \theta \approx \theta\).
When \(\theta \to 0\), \(\cos \theta - 1\) is much smaller than \(\theta\).
Graph Description: Two coordinate plane graphs showing trigonometric behavior. The top graph displays the periodic sine function (blue wave) with a tangent line (purple) at the origin demonstrating the slope relationship \(\frac{\sin \theta}{\theta} \approx 1\) for small angles. The bottom graph shows the cosine function (blue wave) and the function \(\cos \theta - 1\) (labeled on the graph), illustrating how \(\cos \theta - 1\) approaches zero faster than \(\theta\) as \(\theta\) approaches zero.
Solution:
We know that \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\). To use this, suppose \(\theta = 2x\), so as \(x \to 0\), we have \(\theta \to 0\).
\[\frac{\sin(2x)}{x} = 2 \cdot \frac{\sin(2x)}{2x}\] \[\lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} 2 \cdot \frac{\sin(2x)}{2x} = 2 \cdot 1 = 2\]Solution:
Suppose \(\theta = 3x\), so as \(x \to 0\), we have \(\theta \to 0\).
\[\lim_{x \to 0} \frac{x}{\sin(3x)} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{3x}{\sin(3x)}\] \[= \frac{1}{3} \cdot \lim_{\theta \to 0} \frac{\theta}{\sin \theta} = \frac{1}{3} \cdot \frac{1}{1} = \frac{1}{3}\]Solution:
From the last two examples: \(\lim_{x \to 0} \frac{\sin(2x)}{x} = 2\) and \(\lim_{x \to 0} \frac{x}{\sin(3x)} = \frac{1}{3}\)
\[\lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)} = \lim_{x \to 0} \frac{\sin(2x)}{x} \cdot \frac{x}{\sin(3x)} = 2 \cdot \frac{1}{3} = \frac{2}{3}\]General Formula:
For constants \(m\) and \(n\) where \(m, n \neq 0\):
\[\lim_{x \to 0} \frac{\sin(mx)}{\sin(nx)} = \frac{m}{n}\]This can be derived by writing:
\[\lim_{x \to 0} \frac{\sin(mx)}{\sin(nx)} = \lim_{x \to 0} \frac{\sin(mx)}{x} \cdot \frac{x}{\sin(nx)} = \lim_{x \to 0} m \cdot \frac{\sin(mx)}{mx} \cdot \frac{nx}{n} \cdot \frac{1}{\sin(nx)} = \frac{m}{n}\]Proof that \(\frac{d}{dx}(\sin x) = \cos x\):
Using the limit definition of the derivative:
\[\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}\]Using the trigonometric identity \(\sin(A + B) = \sin A \cos B + \cos A \sin B\):
\[\sin(x + h) = \sin x \cos h + \cos x \sin h\]Substituting:
\[= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\] \[= \lim_{h \to 0} \frac{\sin x(\cos h - 1) + \cos x \sin h}{h}\] \[= \lim_{h \to 0} \left[\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\right]\]Using our fundamental limits: \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\) and \(\lim_{h \to 0} \frac{\sin h}{h} = 1\)
\[= \sin x \cdot 0 + \cos x \cdot 1 = \cos x\]Graph Description: A coordinate plane showing the sine function \(f(x) = \sin x\) (blue curve) and its derivative \(f'(x) = \cos x\) (red curve). The visual illustrates that where the sine curve has positive slope (increasing), the cosine curve is positive; where sine has negative slope (decreasing), cosine is negative; and where sine has horizontal tangents (at maximum and minimum points), cosine crosses the x-axis (equals zero).
Proof that \(\frac{d}{dx}(\cos x) = -\sin x\):
Using the limit definition of the derivative:
\[\frac{d}{dx}(\cos x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}\]Using the trigonometric identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\):
\[\cos(x + h) = \cos x \cos h - \sin x \sin h\]Substituting:
\[= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}\] \[= \lim_{h \to 0} \frac{\cos x(\cos h - 1) - \sin x \sin h}{h}\] \[= \lim_{h \to 0} \left[\cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h}\right]\]Using our fundamental limits:
\[= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x\]Proof that \(\frac{d}{dx}(\tan x) = \sec^2 x\):
We use the fact that \(\tan x = \frac{\sin x}{\cos x}\) and apply the Quotient Rule:
\[\frac{d}{dx}(\tan x) = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\frac{d}{dx}(\sin x) \cdot \cos x - \sin x \cdot \frac{d}{dx}(\cos x)}{(\cos x)^2}\] \[= \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}\] \[= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x\]Similarly: Using the identity \(\cot x = \frac{\cos x}{\sin x}\) and the Quotient Rule:
\[\frac{d}{dx}(\cot x) = -\csc^2 x\]Proof that \(\frac{d}{dx}(\sec x) = \sec x \tan x\):
We use the fact that \(\sec x = \frac{1}{\cos x}\) and apply the Quotient Rule:
\[\frac{d}{dx}(\sec x) = \frac{d}{dx}\left[\frac{1}{\cos x}\right] = \frac{\frac{d}{dx}(1) \cdot \cos x - 1 \cdot \frac{d}{dx}(\cos x)}{(\cos x)^2}\] \[= \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x}\] \[= \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \cdot \sec x = \sec x \tan x\]Similarly: Using the identity \(\csc x = \frac{1}{\sin x}\) and the Quotient Rule:
\[\frac{d}{dx}(\csc x) = -\csc x \cot x\]Complete List:
\[\frac{d}{dx}(\sin x) = \cos x\] \[\frac{d}{dx}(\cos x) = -\sin x\] \[\frac{d}{dx}(\tan x) = \sec^2 x\] \[\frac{d}{dx}(\sec x) = \sec x \tan x\] \[\frac{d}{dx}(\cot x) = -\csc^2 x\] \[\frac{d}{dx}(\csc x) = -\csc x \cot x\]Solution:
First, simplify the expression using trigonometric identities:
\[\frac{\tan x - 1}{\sec x} = \frac{\frac{\sin x}{\cos x} - 1}{\frac{1}{\cos x}} = (\sin x - \cos x) \cdot \cos x = \sin x \cos x - \cos^2 x\]However, we can also use the Quotient Rule directly. Let \(f(x) = \tan x - 1\) and \(g(x) = \sec x\).
Then \(f'(x) = \sec^2 x\) and \(g'(x) = \sec x \tan x\).
\[\frac{d}{dx}\left[\frac{\tan x - 1}{\sec x}\right] = \frac{(\sec^2 x)(\sec x) - (\tan x - 1)(\sec x \tan x)}{(\sec x)^2}\] \[= \frac{\sec^3 x - \sec x \tan^2 x + \sec x \tan x}{\sec^2 x}\] \[= \frac{\sec x(\sec^2 x - \tan^2 x + \tan x)}{\sec^2 x}\]Using the identity \(\sec^2 x - \tan^2 x = 1\):
\[= \frac{\sec x(1 + \tan x)}{\sec^2 x} = \frac{1 + \tan x}{\sec x} = \cos x + \sin x\]Solution:
First simplify by rationalizing:
\[\frac{x - a}{\sqrt{x} - \sqrt{a}} = \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{\sqrt{x} - \sqrt{a}} = \sqrt{x} + \sqrt{a}\]Taking the derivative:
\[\frac{d}{dx}[\sqrt{x} + \sqrt{a}] = \frac{d}{dx}[x^{1/2} + a^{1/2}] = \frac{1}{2}x^{-1/2} + 0 = \frac{1}{2\sqrt{x}}\]Solution:
First, simplify \(\cos x \tan x = \cos x \cdot \frac{\sin x}{\cos x} = \sin x\).
So we need to find \(\frac{d}{dx}[\sin x \cdot e^x \sec x]\).
Using the Product Rule (treating this as a product of two functions):
\[= \frac{d}{dx}(\sin x) \cdot (e^x \sec x) + \sin x \cdot \frac{d}{dx}(e^x \sec x)\]For the second term, we need the Product Rule again:
\[\frac{d}{dx}(e^x \sec x) = \frac{d}{dx}(e^x) \cdot \sec x + e^x \cdot \frac{d}{dx}(\sec x)\] \[= e^x \sec x + e^x \sec x \tan x\]Putting it all together:
\[= \cos x \cdot e^x \sec x + \sin x \cdot (e^x \sec x + e^x \sec x \tan x)\] \[= \cos x \cdot e^x \sec x - e^x \sec x - e^x \sec x \tan x\]