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Chapter 8: Problem 1

$$\text { Is } \theta=\pi / 2 \text { a solution of } 2 \cos ^{2} \theta-3 \cos \theta=0 ?$$

Short Answer

Expert verified
\(\theta=\frac{\pi}{2}\) is a solution of the equation.

Step by step solution

01

Plug in the value

To determine if \(\theta = \frac{\pi}{2}\) is a solution, we will substitute \(\theta = \frac{\pi}{2}\) into the given equation \(2 \cos^2 \theta - 3 \cos \theta = 0\). This becomes \(2 \cos^2\left(\frac{\pi}{2}\right) - 3 \cos\left(\frac{\pi}{2}\right) = 0\).
02

Calculate cosine at \(\frac{\pi}{2}\)

The cosine of \(\frac{\pi}{2}\) is \(0\), so \(\cos\left(\frac{\pi}{2}\right) = 0\). Substitute this into the equation: \(2(0)^2 - 3(0) = 0\).
03

Simplify the equation

Simplify the equation: \(2 \times 0 - 3 \times 0 = 0\). This simplifies to \(0 = 0\).
04

Final Step: Verify the result

Since we have simplified and found that both sides of the equation equal \(0\), \(\theta = \frac{\pi}{2}\) is indeed a solution.

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Key Concepts

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

Cosine Function
The cosine function is one of the fundamental trigonometric functions, often appearing in various mathematical and real-world applications. Its general form is the cosine of an angle \(\theta\), written as \(\cos(\theta)\). This function measures the adjacent side of a right triangle over its hypotenuse in a right-angled triangle.
  • The cosine function is periodic, with a period of \(2\pi\), meaning its values repeat every \(2\pi\) radians.
  • It ranges from -1 to 1, taking these values at critical points within its cycle.
  • At \(\theta = \frac{\pi}{2}\), the cosine value is \(0\).

The property that \(\cos\left(\frac{\pi}{2}\right) = 0\) is crucial in solving the exercise, where substituting this value simplifies the equation to \(0 = 0\). Understanding the behavior of the cosine function helps in analyzing its role in various trigonometric identities and equations.
Angle Substitution
Angle substitution is a method used in solving trigonometric equations by replacing a trigonometric function with a specific angle to simplify calculations. It allows us to check the validity of an angle as a potential solution.
  • In this exercise, angle substitution involves replacing \(\theta\) with \(\frac{\pi}{2}\).
  • This substitution is applied directly to the trigonometric equation \(2 \cos^2 \theta - 3 \cos \theta = 0\).

The value of \(\cos\left(\frac{\pi}{2}\right)\) becomes zero through this process, making it straightforward to substitute back into the equation, simplifying it considerably. This method is helpful in verifying potential solutions of trigonometric equations by reducing complex expressions to simpler forms for easy evaluation.
Solution Verification
Solution verification is a critical step in ensuring a proposed solution actually satisfies the given equation. It involves substituting the solution back into the original equation and confirming both sides equate.
  • For the given equation, after substituting \(\theta = \frac{\pi}{2}\), the expression becomes \(2 \times 0 - 3 \times 0 = 0\).
  • Simplifying these terms results in \(0 = 0\).
  • The equality confirms that no discrepancies exist, verifying that \(\theta = \frac{\pi}{2}\) is indeed a solution.

This practice ensures accuracy and reduces error, providing confidence that the solution holds true in the context of the given problem. Keep in mind, verification is a necessary component in solving trigonometric problems and validating results effectively.

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Most popular questions from this chapter

Use the given information to express \(\sin 2 \theta\) and \(\cos 2 \theta\) in terms of \(x\). $$x+1=3 \sin \theta \quad\left(\frac{\pi}{2}<\theta<\pi\right)$$

Evaluate the given expressions without using a calculator or tables. $$\csc \left[\sin ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1}\left(\frac{1}{2}\right)\right]$$

Calculation of \(\sin 18^{\circ}, \cos 18^{\circ},\) and \(\sin 3^{\circ}\) (a) Prove that \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\) (b) Supply a reason for each statement. (i) \(\sin 36^{\circ}=\cos 54^{\circ}\) (ii) \(2 \sin 18^{\circ} \cos 18^{\circ}=4 \cos ^{3} 18^{\circ}-3 \cos 18^{\circ}\) (iii) \(2 \sin 18^{\circ}=4 \cos ^{2} 18^{\circ}-3\) (c) In equation (iii), replace \(\cos ^{2} 18^{\circ}\) by \(1-\sin ^{2} 18^{\circ}\) and then solve the resulting equation for \(\sin 18^{\circ} .\) Thus show that \(\sin 18_{i}=\frac{1}{4}(\sqrt{5}-1)\) (d) Show that \(\cos 18_{i}=\frac{1}{4} \sqrt{10+2 \sqrt{5}}\) (e) Show that \(\sin 3^{\circ}\) is equal to \(\frac{1}{16}[(\sqrt{5}-1)(\sqrt{6}+\sqrt{2})-2(\sqrt{3}-1) \sqrt{5+\sqrt{5}}]\) (f) Use your calculator to check the results in parts (c), (d), and (e).

Prove each of the following double-angle formulas. Hint: As in the text, replace \(2 \theta\) with \(\theta+\theta,\) and use an appropriate addition formula. (a) \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\) (b) \(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\)

Prove that the given equations are identities. $$\sin ^{4} \theta=\frac{3-4 \cos 2 \theta+\cos 4 \theta}{8}$$
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