Question

\(f(x)=\sin x, g(x)=\cos x, h(x)=2 x,\) and \(p(x)=\frac{x}{2} .\) Find the value of each of the following: $$ (h \circ f)\left(\frac{\pi}{6}\right) $$

Short answer

1

Step-by-step solution

Understand the Notation

The notation \(h \circ f\) indicates composition of functions. Specifically, \(h \circ f(x) \) means \(h(f(x))\). You need to apply \(f(x)\) first and then use the result as the input for \(h(x)\).

Apply Function f

Identify \(f(x) = \sin x\) and find \(f\left(\frac{\pi}{6}\right)\). Compute \[ f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

Apply Function h

Use the result from Step 2 as the input for \(h(x)\). Recall \(h(x) = 2x\). Now compute \[ h\left(f\left(\frac{\pi}{6}\right)\right) = h\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} = 1 \]

Final Answer

The value of \(h\left(f\left(\frac{\pi}{6}\right)\right)\) is 1.

Key concepts

Trigonometric Functions

Trigonometric functions are fundamental in trigonometry, a branch of mathematics dealing with the relationships between the angles and sides of triangles. These functions include the sine (sin), cosine (cos), and tangent (tan), among others. Understanding these functions is crucial in various fields such as physics, engineering, and even economics. In this exercise, we're particularly dealing with the sine and cosine functions. These functions map an angle (usually measured in radians) to a specific ratio. For instance, the sine of an angle gives us the ratio of the length of the opposite side to the hypotenuse in a right triangle. This property of trigonometric functions makes them invaluable for solving problems involving periodic phenomena, such as sound and light waves.

Sine Function

The sine function, denoted as \(\text{sin}(x)\), is one of the primary trigonometric functions. It is defined in the context of a right triangle or on the unit circle. When using the unit circle, \(sin(x)\) represents the y-coordinate of the point where the terminal side of the angle intersects the circle. For example, \(sin\big(\frac{\pi}{6}\big) = \frac{1}{2}\). This means that for an angle of \frac{\pi}{6}\ radians (or 30 degrees) in the unit circle, the sine value is \frac{1}{2}\. In the original exercise, we needed to find \(\text{sin}\big(\frac{\pi}{6}\big)\), which was crucial for later steps. By understanding that sine translates angles to specific ratios, we can apply this concept to various mathematical and real-world problems.

Function Evaluation

Function evaluation involves substituting a specific input into a function to find the corresponding output. In this problem, we worked with function composition, which means evaluating one function and then using its output as the input for another function. For instance, we first evaluated the function \(f(x) = \sin x\) at \(\frac{\pi}{6}\), giving us \(\frac{1}{2}\). After getting this result, we used it as the input for the function \(h(x) = 2x\). The step-by-step approach to function evaluation allows us to tackle complex problems by breaking them down into simpler, more manageable parts.

Algebraic Operations

Algebraic operations like addition, subtraction, multiplication, and division are the backbone of manipulating functions. In this exercise, we utilized multiplication in the final step to find our answer. Specifically, after determining that \(\text{sin}\big(\frac{\pi}{6}\big) = \frac{1}{2}\), we used this result as input for the function \(h(x) = 2x\). To evaluate \(h\big(\frac{1}{2}\big)\), we multiplied \(\frac{1}{2}\) by 2, yielding 1. Understanding these basic operations is essential for performing more advanced calculations, whether they involve single functions or compositions of multiple functions.