Finding constants in a harmonic form equation

MathematicsTrigonometric FunctionsMediumSTEM

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6 A curve has equation $$y = a \sin x + b \cos x$$ where $a$ and $b$ are constants. The maximum value of $y$ is $4$ and the curve passes through the point $\left(\frac{\pi}{3}, 2\sqrt{3}\right)$ as shown in the diagram. Find the exact values of $a$ and $b$.

This question includes visual content: A Cartesian coordinate system showing a sinusoidal wave labeled y = a sin x + b cos x. A specific point on the curve is marked with the coordinates (pi/3, 2sqrt(3)). The graph shows the function passing through the y-axis, reaching a peak, then dropping through the marked point to a local minimum.

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Step by Step Written Solution

1
Step 1

Hi Ivam, let's solve this problem involving a trigonometric curve and find the exact values of the constants a and b.

Problem Analysis

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Step 2

We are given the equation y equals a sin x plus b cos x. We also know that the maximum value of y is four, and the curve passes through the point pi over three, two root three.

$$y = a \sin x + b \cos x$$

1. Maximum value of $y = 4$

2. Point $(\frac{\pi}{3}, 2\sqrt{3})$ lies on the curve

3
Step 3

Let's start by using the information about the maximum value. A linear combination of sine and cosine like this can be rewritten as a single harmonic term.

Using the Maximum Value

$$a \sin x + b \cos x = R \sin(x + \alpha)$$
4
Step 4

The amplitude R of this combined wave is equal to the square root of a squared plus b squared. Since the maximum value of the sine function is one, the maximum value of y is simply R.

$$R = \sqrt{a^2 + b^2}$$
$$y_{\text{max}} = R = 4$$
5
Step 5

By squaring both sides, we get our first equation: a squared plus b squared equals sixteen.

$$a^2 + b^2 = 16 \quad \text{--- (1)}$$
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Step 6

Now, let's use the given point to create a second equation. We substitute x equals pi over three and y equals two root three into our original equation.

Using the Point Subsitution

$$y = a \sin x + b \cos x$$
$$2\sqrt{3} = a \sin\left(\frac{\pi}{3}\right) + b \cos\left(\frac{\pi}{3}\right)$$
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Step 7

Recall the exact values for sine and cosine at pi over three. Sine of pi over three is root three over two, and cosine of pi over three is one half.

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Step 8

To simplify, let's multiply the entire equation by two to clear the denominators.

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Step 9

We can now express b in terms of a, which gives us b equals four root three minus a root three. This is our second equation.

$$b = \sqrt{3}(4 - a) \quad \text{--- (2)}$$
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Step 10

Now we can solve for a by substituting equation two into equation one.

Solving for a and b

$$a^2 + b^2 = 16$$
$$a^2 + [\sqrt{3}(4 - a)]^2 = 16$$

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About This Question

Subject
Mathematics
Topic
Trigonometric Functions
Difficulty
Medium
Exam
STEM
Question Type
Open Ended

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