Problem 64 - Contemporary Precalculus

Question

(a) Find two numbers \(c\) and \(d\) such that $$ \sin (c+d) \neq \sin c+\sin d $$ (b) Find two numbers \(c\) and \(d\) such that $$ \cos (c+d) \neq \cos c+\cos d $$ (TABLE CAN'T COPY)

Step-by-Step Solution

Answer

Question: Find two numbers c and d such that the sum-to-product trigonometric identities do not hold for both sine and cosine functions. Answer: The numbers c = π/3 and d = π/6 satisfy the required inequalities for both sine and cosine functions.

Step 1: Find two numbers c and d for sine function

We are looking for c and d such that \(\sin (c+d) \neq \sin c + \sin d\). We can test this inequality with some angles. Let's start with \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\).

Step 2: Compute the sine of the sum of the angles

Calculate \(\sin (c+d)\) for the selected angles \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\): $$ \sin (c+d) = \sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right) = \sin \frac{\pi}{2} = 1 $$

Step 3: Compute the sum of the sines

Now we need to calculate the sum of the sines of the angles: \(\sin c + \sin d = \sin \frac{\pi}{3} + \sin \frac{\pi}{6}\): $$ \sin c + \sin d = \sin \frac{\pi}{3} + \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2} $$

Step 4: Check the inequality

Compare the results: $$ \sin (c+d) \neq \sin c + \sin d \Rightarrow 1 \neq \frac{\sqrt{3}}{2} + \frac{1}{2} $$ The inequality holds, so we have found two numbers c and d that satisfy part (a).

Step 5: Find two numbers c and d for cosine function

We are looking for c and d such that \(\cos (c+d) \neq \cos c + \cos d\). We can use the same angles we used for sine function: \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\).

Step 6: Compute the cosine of the sum of the angles

Calculate \(\cos (c+d)\) for the selected angles \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\): $$ \cos (c+d) = \cos \left(\frac{\pi}{3}+\frac{\pi}{6}\right) = \cos \frac{\pi}{2} = 0 $$

Step 7: Compute the sum of the cosines

Now we need to calculate the sum of the cosines of the angles: \(\cos c + \cos d = \cos \frac{\pi}{3} + \cos \frac{\pi}{6}\): $$ \cos c + \cos d = \cos \frac{\pi}{3} + \cos \frac{\pi}{6} = \frac{1}{2} + \frac{\sqrt{3}}{2} $$

Step 8: Check the inequality

Compare the results: $$ \cos (c+d) \neq \cos c + \cos d \Rightarrow 0 \neq \frac{1}{2} + \frac{\sqrt{3}}{2} $$ The inequality holds, so we have found two numbers c and d that satisfy part (b). In conclusion, we found the numbers \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\) that satisfy the required inequalities for both sine and cosine functions.

Key Concepts

Sine and Cosine Functions

Trigonometry is the branch of mathematics that studies relationships between side lengths and angles in triangles. Two fundamental trigonometric functions are sine (sin) and cosine (cos). They are crucial for dealing with right-angled triangles and modeling periodic phenomena.

The sine function describes the ratio of the length of the opposite side to the length of the hypotenuse in a right-angled triangle, while the cosine function describes the ratio of the length of the adjacent side to the hypotenuse. These ratios are defined for any angle, not just angles in right triangles, via the unit circle. In the unit circle, the angle measure is the arc length around the circle, and the coordinates of a point on the circle are \(\cos\theta, \sin\theta\).

For any angle \(\theta\), it's important to note that \(-1 \leq \sin\theta \leq 1\) and \(-1 \leq \cos\theta \leq 1\), which is derived from the unit circle. A common mistake is to assume that the functions behave linearly, as in the exercise, where it is incorrectly presumed that \(\sin(c+d)\) could equal \(\sin c + \sin d\) or \(\cos(c+d)\) could equal \(\cos c + \cos d\). This exercise shows that is not the case.

Angle Addition Formulas

In trigonometry, angle addition formulas are useful to simplify the computation of the sine or cosine of a sum or difference of angles. The angle addition formulas for sine and cosine are:

• \(\sin(c + d) = \sin c \cdot \cos d + \cos c \cdot \sin d\)
• \(\cos(c + d) = \cos c \cdot \cos d - \sin c \cdot \sin d\)

These formulas tell us that, unlike algebraic sums, we can't just add the sines or cosines of two angles to get the sine or cosine of their sum. In the given exercise, the provided angle values of \(c = \frac{\pi}{3}\) and \(d = \frac{\pi}{6}\) demonstrate this principle.

When we calculate \(\sin(c+d)\), we actually get a value of 1, whereas the sum of \(\sin c + \sin d\) yields \(\frac{\sqrt{3}}{2} + \frac{1}{2}\), which clearly does not equal 1. Similarly, for cosine, the formula prevents us from simply adding \(\cos c\) and \(\cos d\) to find \(\cos(c+d)\), as illustrated by the different results when we do the computations according to the angle addition formulas.

Mathematical Inequalities

Mathematical inequalities express the relationship between two values indicating that they are not equal and that one is greater or less than the other. Inequalities are not only fundamental in algebra but also play a significant role in trigonometry and calculus. They enable us to compare values and establish the range of possible solutions for equations.

In the context of trigonometric functions, the values for sine and cosine are bounded, meaning that they lie within a certain range, as discussed earlier. The inequalities from the exercise show that because of these bounds and the properties of trigonometric functions, misunderstanding the addition of angles can result in incorrect equations.

The provided solution uses evaluation and comparison to prove that the inequalities hold. By comparing the values obtained from applying the correct angle addition formulas to the incorrect assumption of linear addition, the inequality is established—demonstrating that \(\sin (c+d)\) is not equal to \(\sin c + \sin d\) and \(\cos (c+d)\) is not equal to \(\cos c + \cos d\). This illustrates the practical importance of understanding and applying inequalities in trigonometry.