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Chapter 5: Problem 19

Write each expression as a product of sines and/or cosines. $$ \sin \left(\frac{x}{2}\right)-\sin \left(\frac{5 x}{2}\right) $$

Short Answer

Expert verified
The expression is \(-2 \cos(\frac{3x}{2}) \sin(\frac{x}{2})\)."

Step by step solution

01

Use the Sine Difference Identity

We start by using the sine difference identity: \( \sin A - \sin B = 2 \cos \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right) \). Here, let \( A = \frac{x}{2} \) and \( B = \frac{5x}{2} \).
02

Calculate \( \frac{A + B}{2} \) and \( \frac{A - B}{2} \)

First, calculate \( \frac{A + B}{2} \):\[ \frac{\frac{x}{2} + \frac{5x}{2}}{2} = \frac{6x}{4} = \frac{3x}{2} \]Next, calculate \( \frac{A - B}{2} \):\[ \frac{\frac{x}{2} - \frac{5x}{2}}{2} = \frac{-4x}{4} = -\frac{x}{2} \]
03

Substitute Back Into the Identity

Substitute these results into the sine difference identity:\[ \sin \left(\frac{x}{2}\right) - \sin \left(\frac{5x}{2}\right) = 2 \cos \left(\frac{3x}{2}\right) \sin \left(-\frac{x}{2}\right) \]
04

Simplify Using the Odd Function Property

Since sine is an odd function, \( \sin(-x) = -\sin(x) \). Therefore, \( \sin\left(-\frac{x}{2}\right) = -\sin\left(\frac{x}{2}\right) \).Substitute this into the expression:\[ 2 \cos \left(\frac{3x}{2}\right) \sin \left(-\frac{x}{2}\right) = 2 \cos \left(\frac{3x}{2}\right)(-\sin \left(\frac{x}{2}\right)) \]
05

Final Expression as a Product

Finally, multiply the terms: \[ 2 \cos \left(\frac{3x}{2}\right)(-\sin \left(\frac{x}{2}\right)) = -2 \cos \left(\frac{3x}{2}\right) \sin \left(\frac{x}{2}\right) \]This is the expression written as a product of sine and cosine.

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Key Concepts

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

Sine Difference Identity
The Sine Difference Identity is a useful trigonometric tool that allows us to express the difference of two sine values as a product involving both sine and cosine functions. This identity states that for any angles \( A \) and \( B \),
  • \( \sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \)
This formula is particularly helpful in simplifying expressions where two sine terms are subtracted from each other. By converting these terms into a sine and cosine product, we can often solve or simplify equations more easily.
In the exercise, we used this identity with \( A = \frac{x}{2} \) and \( B = \frac{5x}{2} \), allowing us to transform the expression into a more manageable form. Understanding this identity is key for solving many trigonometric problems that involve sine differences.
Cosine Function
The cosine function, \( \cos \theta \), is a fundamental trigonometric function defined as the x-coordinate of a point on the unit circle at an angle \( \theta \) from the positive x-axis. It is important to develop a clear understanding of this function as it frequently appears in trigonometric identities and equations.
  • The cosine function is periodic, with a period of \( 2\pi \), meaning that its values repeat every \( 2\pi \) radians.
  • Cosine values range between -1 and 1, inclusive, which represent the maximum and minimum points of the function.
In our exercise, the cosine function appears in the simplified expression as \( 2 \cos \left(\frac{3x}{2}\right) \). Understanding the role of cosine in trigonometric identities is essential for solving problems that involve modifying or decomposing trigonometric functions.
Being able to visualize or graph the cosine function can also provide insight into the properties and behavior of trigonometric identities.
Odd Function Property
The concept of odd functions is another crucial aspect of trigonometric identities. Odd functions have the property that their graph is symmetrical with respect to the origin. Mathematically, this means if a function \( f \) is odd, then \( f(-x) = -f(x) \).
The sine function, \( \sin(x) \), is a perfect example of an odd function. This property allows us to simplify expressions involving negative angles. In the exercise, it was utilized to transform \( \sin(-x) \) into \(-\sin(x)\), simplifying the expression further.
  • Understanding this symmetry is crucial because it allows us to rewrite and simplify expressions with ease.
  • It helps in evaluating trigonometric functions where angles are represented as negative values.
Recognizing and applying the odd function property in trigonometric expressions simplifies many problems, making solution strategies more straightforward and efficient.

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Most popular questions from this chapter

For Exercises 69-72, refer to the following: We cannot prove that an equation is an identity using technology, but we can use technology as a first step to see whether or not the equation seems to be an identity. Using a graphing calculator, plot \(Y_{1}=(2 x)-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{5}}{5 !}\) and \(Y_{2}=\sin (2 x)\) for \(x\) ranging \([-1,1]\). Is \(Y_{1}\) a good approximation to \(Y_{2}\) ?

In Exercises \(57-60\), determine whether each statement is true or false. \(\cos (4 A)-\cos (2 A)=\cos (2 A)\)

If \(\sin x=\frac{1}{3}\), find \(\tan (2 x)\) given \(\cos x<0\). Solution: Use the quotient identity. \(\quad \tan (2 x)=\frac{\sin (2 x)}{\cos x}\) Use the double-angle formula for the sine function. \(\tan (2 x)=\frac{2 \sin x \cos x}{\cos x}\) formula for the sine Cancel the common cosine factors. \(\begin{array}{ll}\text { cosine factors. } & \tan (2 x)=2 \sin x \\ \text { Substitute } \sin x=\frac{1}{3} . & \tan (2 x)=\frac{2}{3}\end{array}\) This is incorrect. What mistake was made?

Consider the triangle below, where the vertex angle measures \(\theta\), the equal sides measure \(a\), the height is \(h\), and half the base is \(b\). (In an isosceles triangle, the perpendicular dropped from the vertex angle divides the triangle into two congruent triangles.) The two triangles formed are right triangles. In the right triangles, \(\sin \left(\frac{\theta}{2}\right)=\frac{b}{a}\) and \(\cos \left(\frac{\theta}{2}\right)=\frac{h}{a}\). Multiply each side of each equation by \(a\) to get \(b=a \sin \left(\frac{\theta}{2}\right), h=a \cos \left(\frac{\theta}{2}\right)\). The area of the entire isosceles triangle is \(A=\frac{1}{2}(2 b) h=b h\). Substitute the values for \(b\) and \(h\) into the area formula. Show that the area is equivalent to \(\frac{a^{2}}{2} \sin \theta\).

Verify the identities. $$ \tan ^{2}\left(\frac{x}{2}\right)=\frac{1-\cos x}{1+\cos x} $$
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