In Trigonometry II, we delve deeper into the study of trigonometric functions and their applications. This topic builds on the foundational knowledge of trigonometry learned in previous classes and introduces more advanced concepts that are crucial for solving complex problems involving angles and triangles.
Trigonometric ratios are defined based on the relationships between the sides of a right-angled triangle. The three primary trigonometric ratios are:
Example: Given a right triangle with an angle $\theta$, where the opposite side is 5 and the hypotenuse is 13, find the sine of $\theta$. Solution: $$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$$
Trigonometric identities are equations involving trigonometric functions that are true for all possible values of the variables. Some common trigonometric identities include:
Example: Prove the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$. Solution: Starting with $\sin^2\theta + \cos^2\theta$, we substitute $\sin^2\theta = 1 - \cos^2\theta$, giving $1 - \cos^2\theta + \cos^2\theta = 1$ which simplifies to $1 = 1$, proving the identity.
Special angles like 30°, 45°, and 60° have exact trigonometric values that are commonly used in calculations. For example:
Example: Find the exact value of $\sin 135°$. Solution: Since 135° is in the second quadrant, the sine function is positive. By using the angle difference formula, we get $\sin 135° = \sin(90° + 45°) = \cos 45° = \frac{\sqrt{2}}{2}$.
Trigonometric equations involve trigonometric functions and are solved by applying trigonometric identities and properties. These equations often have multiple solutions within a specified domain.
Example: Solve the equation $\cos\theta = \sin\theta$ for $0° \leq \theta \leq 360°$. Solution: Using the Pythagorean Identity, we rewrite the equation as $\cos\theta = \sqrt{1 - \cos^2\theta}$. Squaring both sides gives $1 - \cos^2\theta = 1 - \sin^2\theta$, leading to $\sin^2\theta - \cos^2\theta = 0$. Factoring gives $(\sin\theta + \cos\theta)(\sin\theta - \cos\theta) = 0$, so $\theta = 45°$ or $\theta = 135°$.
Solution: $$\tan 60° = \frac{\sin 60°}{\cos 60°} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$
Solution: Starting with $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we know that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ by definition.
Solution: Since $\sin\theta = \cos\theta$ is true for $\theta = 45°$, the solution is $\theta = 45°$.
Solution: Since 120° is in the second quadrant, $\cos 120° = -\cos(180° - 120°) = -\cos 60° = -\frac{1}{2}$.
Solution: Since 150° is in the second quadrant, $\sin 150° = \sin(180° - 30°) = \sin 30° = \frac{1}{2}$.
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