Right Triangle Diagram: The handwritten notes show a right triangle with angle θ, where the sides are labeled as hypotenuse, opposite, and adjacent, illustrating the foundation for trigonometric ratios.
Basic Trigonometric Functions: For angle \(\theta\) in a right triangle:
Reciprocal Functions:
Periodicity: All trigonometric functions are periodic: \(f(\theta) = f(\theta + p)\) for some period \(p\).
Unit Circle Diagram: The handwritten notes show a unit circle with different angles marked in multiple quadrants, illustrating positive (counterclockwise) and negative (clockwise) angle measurement from the positive x-axis.
Angle Measurement:
Unit Circle Definitions:
Signs in Different Quadrants:
Radian Measure: Radian is the conventional unit in mathematics
Circle Arc Diagram: The handwritten notes show a circle with radius r and arc length s, illustrating the relationship between angle measure θ, radius, and arc length.
Arc Length Formula: \(s = r\theta\) (where \(\theta\) is in radians)
Definition: 1 radian is the angle measure when arc length equals the radius.
Sine Wave Graph: The handwritten notes show a complete sine wave with amplitude, period, and key points marked, illustrating the standard sine function y = sin θ.
General Sine Function:
\[f(\theta) = a\sin(b(\theta - c)) + d\]Sine Function Restriction: The handwritten notes show the sine function with the domain restricted to [-π/2, π/2] highlighted, demonstrating how to make the function one-to-one for the inverse to exist.
Inverse Sine: \(\arcsin(x) = \sin^{-1}(x)\)
If \(\sin^{-1}(x) = \theta\), then \(\sin \theta = x\)
Domain: \([-1, 1]\)
Range: \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Restriction for Inverse: Restrict domain to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) ⟹ Range: \([-1, 1]\)
This makes the sine function one-to-one, so the inverse function exists.
Evaluate \(\sin^{-1}\left(\frac{1}{2}\right)\):
Solution:
Let \(\sin^{-1}\left(\frac{1}{2}\right) = \theta\), so \(\sin \theta = \frac{1}{2}\)
\(\theta\) is in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\frac{\pi}{6}\) is in the range,
\(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\)
Cosine Function Restriction: The handwritten notes show the cosine function with domain restricted to [0, π], making it decreasing and one-to-one for the inverse to exist.
Inverse Cosine: \(\arccos(x) = \cos^{-1}(x)\)
If \(\cos^{-1}(x) = \theta\), then \(\cos \theta = x\)
Domain: \([-1, 1]\)
Range: \([0, \pi]\)
Evaluate \(\cos\left(\cos^{-1}\left(\frac{4\pi}{3}\right)\right)\):
Unit Circle Reference: The handwritten notes show a unit circle with angle 4π/3 marked, along with its reference angle and coordinates, helping to find the cosine value.
Solution:
Find \(\theta\) in \([0, \pi]\) such that \(\cos\left(\cos^{-1}\left(\frac{4\pi}{3}\right)\right) = \theta\)
\(\cos \theta = \cos\left(\frac{4\pi}{3}\right)\)
\(\frac{4\pi}{3} = \pi + \frac{\pi}{3}\), so \(\cos\left(\frac{4\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}\)
\(\theta = \frac{2\pi}{3}\) (since this is in \([0, \pi]\) and \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\))
Tangent Function Restriction: The handwritten notes show the tangent function with domain restricted to (-π/2, π/2), avoiding the vertical asymptotes and making it one-to-one.
Inverse Tangent: \(\arctan(x) = \tan^{-1}(x)\)
If \(\tan^{-1}(x) = \theta\), then \(\tan \theta = x\)
Domain: \((-\infty, \infty)\)
Range: \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Find \(\cos(\tan^{-1}(x))\):
Right Triangle Construction: The handwritten notes show a right triangle constructed to represent tan⁻¹(x), with opposite side x, adjacent side 1, and hypotenuse √(1+x²).
Solution:
Let \(\tan^{-1}(x) = \theta\), so \(\tan \theta = x\)
Find \(\cos \theta\)
From the right triangle: \(\tan \theta = x = \frac{\text{opposite}}{\text{adjacent}}\)
Let opposite = \(x\), adjacent = 1
Then hypotenuse = \(\sqrt{x^2 + 1}\)
Therefore: \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{1 + x^2}}\)
Key Strategy for Inverse Trig Problems: When evaluating expressions involving inverse trigonometric functions, construct a right triangle using the definition of the inverse function, then use the triangle to find other trigonometric values.