Problem 59 - Algebra and Trigonometry

Question

Find the exact value of the following under the given conditions: a. \(\cos (\alpha+\beta)\) b. \(\sin (\alpha+\beta)\) c. \(\tan (\alpha+\beta)\) \(\tan \alpha=-\frac{3}{4}, \alpha\) lies in quadrant II, and \(\cos \beta=\frac{1}{3}, \beta\) lies in quadrant I.

Step-by-Step Solution

Answer

The exact values of the trigonometric functions are: \(\cos(\alpha+\beta) = -\frac{4}{15} + \frac{2\sqrt{2}}{5}\), \(\sin(\alpha+\beta) = -\frac{1}{5} - \frac{8\sqrt{2}}{15}\) and \(\tan(\alpha+\beta) = \frac{-15-24\sqrt{2}}{20+6\sqrt{2}}\)

Step 1: Calculate the other trigonometric functions for α

Given that \(\tan \alpha = -\frac{3}{4}\) and that \(\alpha\) is in the second quadrant, where sine is positive and cosine is negative. From the identity \(\tan \alpha = \frac{\sin\alpha}{\cos\alpha}\), we can represent \(\sin\alpha\) and \(\cos\alpha\) in the following way: \(\sin\alpha=\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}=-\frac{3}{5}\) (as sine is positive in the second quadrant) and \(\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\frac{4}{5}\) (as cosine is negative in the second quadrant)

Step 2: Calculate the other trigonometric functions for β

Given that \(\cos \beta = \frac{1}{3}\) and that \(\beta\) is in the first quadrant, where all trigonometric functions are positive. From the identity \(\cos^2\beta = 1 - \sin^2\beta\), we can represent \(\sin\beta\) in the following way: \(\sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}\), and \(\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{2\sqrt{2}}{3}\cdot\frac{3}{1} = 2\sqrt{2}\).

Step 3: Calculate cos(α+β)

Using the trigonometric identity \(\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\), substituting the found values we get: \(\cos(\alpha+\beta) = -\frac{4}{5}\cdot\frac{1}{3}-(-\frac{3}{5}\cdot\frac{2\sqrt{2}}{3}) = -\frac{4}{15} + \frac{2\sqrt{2}}{5}\).

Step 4: Calculate sin(α+β)

Using the trigonometric identity \(\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\), substituting the found values we get: \(\sin(\alpha+\beta) = -\frac{3}{5}\cdot\frac{1}{3} + -\frac{4}{5}\cdot\frac{2\sqrt{2}}{3} = -\frac{1}{5} - \frac{8\sqrt{2}}{15}\).

Step 5: Calculate tan(α+β)

Using the trigonometric identity \(\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\), substituting the found values we get: \(\tan(\alpha+\beta) = \frac{-\frac{1}{5} - \frac{8\sqrt{2}}{15}}{-\frac{4}{15} + \frac{2\sqrt{2}}{5}}\). Simplifying, we get \(\tan(\alpha+\beta) = \frac{-15-24\sqrt{2}}{20+6\sqrt{2}}\).

Key Concepts

Sum and Difference Formulas

Understanding the sum and difference formulas in trigonometry is crucial for solving problems involving angles. These formulas help to find the sine, cosine, and tangent of the sum or difference of two angles. Here are the key formulas you need to know:

• For cosine: \[\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta\]
• For sine: \[\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta\]
• For tangent: \[\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\]

These identities allow you to calculate trigonometric values without directly measuring angles. In the problem, we used the cosine and sine identities to find \(\cos(\alpha + \beta)\) and \(\sin(\alpha + \beta)\). Notice how angles \(\alpha\) and \(\beta\) contribute their values through these formulas.
They combine information about individual angles to give the result of their sum. This technique simplifies otherwise complex calculations and is widely used in scenarios requiring precise angle measures.

Trigonometric Functions in Quadrants

The position of an angle in the coordinate plane affects the sign of its trigonometric functions. Understanding this concept is key to applying the right trigonometric identities in calculations. Each of the four quadrants shows specific properties:

• Quadrant I: All trigonometric functions are positive.
• Quadrant II: Sine is positive; cosine and tangent are negative.
• Quadrant III: Tangent is positive; sine and cosine are negative.
• Quadrant IV: Cosine is positive; sine and tangent are negative.

For angle \(\alpha\), knowing it lies in the second quadrant helps determine that \(\tan \alpha = -\frac{3}{4}\) makes \(\sin \alpha\) positive while \(\cos \alpha\) is negative.
Likewise, for angle \(\beta\) in the first quadrant, all trigonometric values stay positive. Understanding the trigonometric behavior based on quadrants allows you to derive accurate values quickly, without needing a calculator.
Always begin with understanding the angle's quadrant before proceeding to use identities as it dictates the signs in your calculations.

Exact Values of Trigonometric Functions

Finding the exact values of trigonometric functions means deriving results without approximate decimals, usually using fractions or radicals. This is especially useful when dealing with problems that require precision, like those found in pure mathematics or engineering.
In the exercise, given \(\tan \alpha = -\frac{3}{4}\), you calculated \(\sin \alpha\) and \(\cos \alpha\) by using the identity:
\[\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\]
You can find \(\sin \alpha\) and \(\cos \alpha\) based on their relationships and the initial conditions given by the problem. Similarly, \(\cos \beta = \frac{1}{3}\) allowed for calculating \(\sin \beta\) using the Pythagorean identity:\[\sin^2\beta = 1 - \cos^2\beta\]
These calculations ensure precision in measuring angles—"exact values" typically mean the calculations remain in exact terms like \(-\frac{4}{15}\) or \(\frac{2\sqrt{2}}{3}\). This is critical for not introducing rounding errors, especially in complex chains of computations.