Problem 38 - A Textbook of Differential Calculus

Question

If the fundamental period of function \(f(x)=\sin x+\cos \left(\sqrt{4-a^{2}}\right) x\) is \(4 \pi\), then the value of \(a\) is/are \(\begin{array}{ll}\text { (a) } \frac{\sqrt{15}}{2} & \text { (b) }-\frac{\sqrt{15}}{2}\end{array}\) (c) \(\frac{\sqrt{7}}{2}\) (d) \(-\frac{\sqrt{7}}{2}\)

Step-by-Step Solution

Answer

The values of \( a \) are \( \frac{\sqrt{15}}{2} \) and \( -\frac{\sqrt{15}}{2} \).

Step 1: Identify Period of Sinusoidal Functions

The function given is composed of two parts: \( f(x) = \sin x + \cos(\sqrt{4-a^2}x) \). We know that the period of \( \sin x \) is \( 2\pi \).

Step 2: Determine the Period of Cosine Function

The period of \( \cos(kx) \) is given by \( \frac{2\pi}{|k|} \). In this case, \( k = \sqrt{4-a^2} \), so the period is \( \frac{2\pi}{\sqrt{4-a^2}} \).

Step 3: Find the Least Common Multiple

The overall period of the function \( f(x) \) is the least common multiple (LCM) of the periods of \( \sin x \) (which is \( 2\pi \)) and \( \cos(\sqrt{4-a^2}x) \) (whose period is \( \frac{2\pi}{\sqrt{4-a^2}} \)).

Step 4: Set Up Equation for Fundamental Period

The smallest positive period that satisfies both \( \sin x \) and \( \cos(\sqrt{4-a^2}x) \) is given as \( 4\pi \). Thus, one of them needs to be a divisor of the other to result in this combined period. Let's first assume \( \frac{2\pi}{\sqrt{4-a^2}} \) is a multiple of \( 2\pi \).

Step 5: Solve for \( a \) Using the Period Equation

Setting the equation for fundamental period, we have:\[ \frac{2\pi}{\sqrt{4-a^2}} = \frac{2\pi}{2} \]Solving for \( a \):\[ \sqrt{4-a^2} = \frac{1}{2} \]Squaring both sides, \( 4 - a^2 = \frac{1}{4} \).\[ a^2 = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \].

Step 6: Finalize Value of \( a \)

Now find the value of \( a \):\[ a = \pm \sqrt{\frac{15}{4}} = \pm \frac{\sqrt{15}}{2} \].

Key Concepts

Periodicity

Periodicity is the concept that helps us determine how often a function repeats its values. In the case of trigonometric functions, they are known for their periodic nature, meaning they repeat their values at regular intervals. For instance, the sine and cosine functions both have a standard period of

• For \( \sin x \), the period is \( 2\pi \).
• For \( \cos x \), the period is also \( 2\pi \).

However, when we introduce a transformed version of these functions, such as \( \cos(kx) \), where \( k \) is a scaling factor, the period changes. The new period becomes \( \frac{2\pi}{|k|} \). Knowing how to determine the period of transformed trigonometric functions allows us to solve equations where periodicity plays a crucial role.
In the exercise, the concept of periodicity is essential to find the fundamental period of the function \( f(x)=\sin x+\cos(\sqrt{4-a^{2}}x) \). The fundamental period is defined as the smallest positive interval over which the entire function repeats itself. Understanding periodicity helps in solving the problem by finding the least common multiple of the individual periods of the sine and cosine components.

Trigonometric Functions

Trigonometric functions like sine and cosine are vital in many fields, including physics, engineering, and mathematics. These functions describe relationships between the angles and sides of triangles and are foundational in studying waves and oscillations.
Here are the basic forms:

• \( \sin x \) varies between -1 and 1 and is zero at integer multiples of \( \pi \).
• \( \cos x \) also ranges from -1 to 1 and is zero at odd multiples of \( \frac{\pi}{2} \).

Both \( \sin x \) and \( \cos x \) are periodic, with periods of \( 2\pi \), meaning they repeat every \( 2\pi \) units. The exercise illustrates this by combining these two trigonometric functions with different periods. By considering a specific transformation, such as \( \cos(\sqrt{4-a^2}x) \), the period is altered to \( \frac{2\pi}{\sqrt{4-a^2}} \). These two components, \( \sin x \) and \( \cos(\sqrt{4-a^2}x) \), have their individual periodic behaviors which combine to give the entire function a new fundamental period. This insight allows us to solve the problem by equating the common period \( 4\pi \) with the transformed functions' periods.

Solving Equations

Solving equations often involves finding unknown variables that satisfy a given mathematical condition or conditions. In the context of the exercise, we are tasked with determining the value of \( a \) that allows the function \( f(x)=\sin x+\cos(\sqrt{4-a^2}x) \) to have a fundamental period of \( 4\pi \).
Here is the step-by-step approach to solving this kind of problem:

• First, identify the periods of the individual functions inside \( f(x) \): \( \sin x \) with period \( 2\pi \) and \( \cos(\sqrt{4-a^2}x) \) with period \( \frac{2\pi}{\sqrt{4-a^2}} \).
• Next, determine their Least Common Multiple (LCM), as the overall period of the function is the LCM of its components' periods. The exercise states this LCM should be \( 4\pi \).
• Set up an equation based on this LCM, which leads us to \( \frac{2\pi}{\sqrt{4-a^2}} = \frac{2\pi}{2} \).
• Solve for \( a \). By simplifying, we find \( \sqrt{4-a^2} = \frac{1}{2} \), leading to \( a^2 = \frac{15}{4} \).
• Finally, take the square root of both sides to find \( a = \pm \frac{\sqrt{15}}{2} \).

Through these steps, we can reach the conclusion by calculating carefully and understanding each component of the problem.