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Chapter 3: Problem 1

Write the given expression as a product of two trigonometric functions of different frequencies. \(\cos 9 t-\cos 7 t\)

Short Answer

Expert verified
Question: Rewrite the expression \(\cos 9t -\cos 7t\) as a product of two trigonometric functions with different frequencies. Answer: \(-2 \sin(8t)\sin(t)\)

Step by step solution

01

Recall the sum-to-product formula for cosine functions

The sum-to-product formula for the cosine functions is: \(\cos A -\cos B = -2 \sin{\frac{A+B}{2}} \sin{\frac{A-B}{2}}\) We'll use this formula to rewrite the given expression as a product of two trigonometric functions with different frequencies.
02

Identify A and B in the given expression

In the given expression, \(\cos 9t -\cos 7t\), we can identify A and B as follows: \(A = 9t\) and \(B = 7t\)
03

Apply the sum-to-product formula

Using the sum-to-product formula for cosine functions, we can now rewrite the given expression: \(\cos 9t -\cos 7t = -2 \sin\left(\frac{9t+7t}{2}\right) \sin\left(\frac{9t-7t}{2}\right)\)
04

Simplify the expression

Now, we can simplify the expression further: \(-2 \sin\left(\frac{9t+7t}{2}\right) \sin\left(\frac{9t-7t}{2}\right) = -2 \sin\left(\frac{16t}{2}\right) \sin\left(\frac{2t}{2}\right)\) \(-2 \sin(8t)\sin(t)\) So, the given expression \(\cos 9t -\cos 7t\) can be written as a product of two trigonometric functions of different frequencies: \(-2 \sin(8t)\sin(t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Understanding trigonometric functions is essential to solving many problems in mathematics, especially in trigonometry. Trigonometric functions describe the relationships between the angles and lengths of a right triangle. The primary functions are sine (sin), cosine (cos), and tangent (tan), each with specific meanings in the context of a right triangle.

However, their applications go beyond just triangles; they are periodic and can be used to model waves and oscillations, like sound waves and tides. Each function has a unique graph that repeats at regular intervals, known as the period. The beauty of these functions lies in their interrelated identities, such as sum-to-product, which enables the simplification of complex trigonometric expressions and the solution of equations.
Sine Function
The sine function is one of the fundamental trigonometric functions, defined as the ratio of the length of the opposite side to the length of the hypotenuse in a right-angled triangle. When looking at the unit circle, which is a circle with a radius of 1, the sine of an angle is the y-coordinate of the point where a line at that angle intersects the circle.

Understanding the sine function is crucial for grasping trigonometric identities and solving problems. The sine function is periodic, with a period of \(2\pi\) or 360 degrees, meaning that it repeats its values in regular cycles. This periodic nature makes it incredibly useful in describing cyclic phenomena, such as the one presented in the exercise where it helps in transforming a sum of cosine functions into a product of sine functions.
Trigonometry Problem Solving
Trigonometry problem solving involves strategies to simplify and solve equations or expressions involving trigonometric functions. One of the key problem-solving techniques includes the use of trigonometric identities. These identities are formulas that relate the trigonometric functions and are true for all values of the involved angles.

For example, sum-to-product identities are specifically helpful when dealing with the addition or subtraction of two trigonometric functions, allowing for the conversion into a product of functions. Just as shown in the step-by-step solution, by understanding and applying these identities, complex trigonometry problems become more manageable. Certainly, sum-to-product identities offer a way to rewrite and hence simplify the expressions which lead to either solving the trigonometry equations or simplifying the expression for further analysis.

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Most popular questions from this chapter

In this problem we determine conditions on \(p\) and \(q\) such that \(\mathrm{Eq}\). (i) can be transformed into an equation with constant coefficients by a change of the independent variable, Let \(x=u(t)\) be the new independent variable, with the relation between \(x\) and \(t\) to be specified later. (a) Show that $$ \frac{d y}{d t}=\frac{d x}{d t} \frac{d y}{d x}, \quad \frac{d^{2} y}{d t^{2}}=\left(\frac{d x}{d t}\right)^{2} \frac{d^{2} y}{d x^{2}}+\frac{d^{2} x}{d t^{2}} \frac{d y}{d x} $$ (b) Show that the differential equation (i) becomes $$ \left(\frac{d x}{d t}\right)^{2} \frac{d^{2} y}{d x^{2}}+\left(\frac{d^{2} x}{d t^{2}}+p(t) \frac{d x}{d t}\right) \frac{d y}{d x}+q(t) y=0 $$ (c) In order for Eq. (ii) to have constant coefficients, the coefficients of \(d^{2} y / d x^{2}\) and of \(y\) must be proportional. If \(q(t)>0,\) then we can choose the constant of proportionality to be \(1 ;\) hence $$ x=u(t)=\int[q(t)]^{1 / 2} d t $$ (d) With \(x\) chosen as in part (c) show that the coefficient of \(d y / d x\) in Eq. (ii) is also a constant, provided that the expression $$ \frac{q^{\prime}(t)+2 p(t) q(t)}{2[q(t)]^{3 / 2}} $$ $$ \frac{q^{\prime}(t)+2 p(t) q(t)}{2[q(t)]^{3 / 2}} $$ is a constant. Thus Eq. (i) can be transformed into an equation with constant coefficients by a change of the independent variable, provided that the function \(\left(q^{\prime}+2 p q\right) / q^{3 / 2}\) is a constant. How must this result be modified if \(q(t)<0 ?\)

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

The position of a certain undamped spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+2 u=0, \quad u(0)=0, \quad u^{\prime}(0)=2 $$ (a) Find the solution of this initial value problem. (b) Plot \(u\) versus \(t\) and \(u^{\prime}\) versus \(t\) on the same axes. (c) Plot \(u^{\prime}\) versus \(u ;\) that is, plot \(u(t)\) and \(u^{\prime}(t)\) parametrically with \(t\) as the parameter. This plot is known as a phase plot and the \(u u^{\prime}\) -plane is called the phase plane. Observe that a closed curve in the phase plane corresponds to a periodic solution \(u(t) .\) What is the direction of motion on the phase plot as \(t\) increases?

Assume that the system described by the equation \(m u^{\prime \prime}+\gamma u^{\prime}+k u=0\) is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Hint: Determine all possible values of \(t\) for which \(u=0\).

In each of Problems 1 through 12 find the general solution of the given differential equation. $$ y^{\prime \prime}-2 y^{\prime}-3 y=3 e^{2 x} $$
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