Problem 63 - Precalculus Essentials

Question

Find the exact value of the following under the given conditions: (A) .\(\cos (\alpha+\beta)\) (B). \(\sin (\alpha+\beta)\) (C) \(\tan (\alpha+\beta)\). \(\tan \alpha=\frac{3}{4}, \pi<\alpha<\frac{3 \pi}{2},\) and \(\cos \beta=\frac{1}{4}, \frac{3 \pi}{2}<\beta<2 \pi\)

Step-by-Step Solution

Answer

\(\cos(\alpha + \beta)=-\frac{1}{5}-\frac{3\sqrt{15}}{20}, \sin(\alpha + \beta) = -\frac{3}{20}+\sqrt{15}, \tan(\alpha + \beta) = \frac{-\frac{3}{20}+\sqrt{15}}{-\frac{1}{5}- \frac{3\sqrt{15}}{20}}\)

Step 1: Find \(\sin \alpha\) and \(\cos \alpha\)

Given \(\tan \alpha = \frac{3}{4}\) This means that the opposite side (O) is 3 units and adjacent side (A) is 4 units. It happens to be in third quadrant where cosine is negative and sine is negative.By the Pythagorean theorem, the hypotenuse (H) = \(\sqrt{O^2 + A^2} = 5\) units.Therefore we can find that \(\sin \alpha = -\frac{O}{H} = -\frac{3}{5}\) and \(\cos \alpha = -\frac{A}{H} = -\frac{4}{5}\)

Step 2: Determine \(\sin \beta\)

Given \(\cos \beta = \frac{1}{4}\). As \(\beta\) is in the fourth quadrant where sine is negative and cosine is positive, we can calculate \(\sin \beta\).We will use the identity \(\cos^2 \beta + \sin^2 \beta = 1\)Solving for sin:\(\sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{1}{4}\right)^2 = \frac{15}{16}\)Taking the square root gives \(\sin \beta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}\)

Step 3: Evaluate \(\cos(\alpha + \beta), \sin(\alpha + \beta) and \(\tan(\alpha + \beta)\)

Now apply the formulae. \(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\)Substitute the values to get the result:\(\cos(\alpha + \beta) = -\frac{4}{5} \times \frac{1}{4} - -\frac{3}{5} \times -\frac{\sqrt{15}}{4} = -\frac{1}{5} - \frac{3\sqrt{15}}{20}\)Similarly,\(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\)On substituting the values, we get:\(\sin(\alpha + \beta) = -\frac{3}{5} \times \frac{1}{4} + -\frac{4}{5} \times -\frac{\sqrt{15}}{4} = -\frac{3}{20} + \sqrt{15}\)And finally,\(\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{-\frac{3}{20}+\sqrt{15}}{-\frac{1}{5}- \frac{3\sqrt{15}}{20}}\)

Key Concepts

Sum of Angles Formula

Understanding the sum of angles formula is crucial to solving trigonometric equations involving the sum or difference of two angles. When dealing with trigonometric functions such as sine, cosine, and tangent, we apply these specific formulas to simplify the expressions.

• For cosine, the formula is \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \).
• For sine, it's \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \).
• The tangent sum formula, derived from the sine and cosine formulas, is \( \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \).

These equations allow us to find the trigonometric values of the sum or differences of angles simply by knowing the values of the individual angles' trigonometric functions.

Pythagorean Theorem

The Pythagorean theorem is a fundamental concept in mathematics, especially when dealing with right-angled triangles. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
Expressed in a formula, it is \( c^2 = a^2 + b^2 \), where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. In trigonometry, this theorem helps us find the unknown side when we have the lengths of two sides of the triangle, and it is also inherent in the fundamental relationships between the trigonometric functions, like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Trigonometric Functions

Trigonometric functions are the backbone of trigonometry and are essential in understanding the relationships between angles and sides of triangles. The primary trigonometric functions are sine (sin), cosine (cos), and tangent (tan), and they are defined as ratios of sides in a right triangle relative to one of its acute angles.

• \textbf{Sine} of an angle is the ratio of the length of the opposite side to the length of the hypotenuse: \( \sin \theta = \frac{opposite}{hypotenuse} \).
• \textbf{Cosine} is the ratio of the length of the adjacent side to the hypotenuse: \( \cos \theta = \frac{adjacent}{hypotenuse} \).
• \textbf{Tangent} is the ratio of the sine and cosine of the angle, or equivalently, the opposite side over the adjacent side: \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{opposite}{adjacent} \).

These functions enable us to solve various problems in geometry, physics, engineering, and many other fields.

Quadrant Rule for Trigonometry

The quadrant rule is a simple way to determine the sign of trigonometric functions based on the angle's quadrant location on the unit circle. The unit circle is divided into four quadrants, and each trigonometric function - sine, cosine, and tangent - has a specific sign in each quadrant.

• In the \textbf{first quadrant} (0 to \( \frac{\pi}{2} \) radians), all trigonometric functions are positive.
• In the \textbf{second quadrant} (\( \frac{\pi}{2} \) to \( \pi \) radians), sine is positive, but cosine and tangent are negative.
• The \textbf{third quadrant} (\( \pi \) to \( \frac{3\pi}{2} \) radians) sees sine and cosine as negative, while tangent is positive.
• In the \textbf{fourth quadrant} (\( \frac{3\pi}{2} \) to 2\textpi radians), cosine is positive, and sine and tangent are negative.

By knowing where an angle lies on the unit circle, we can easily determine the sign of its trigonometric functions, which is crucial when solving more complex equations.