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Chapter 1: Problem 48

Find the function values in Exercises \(47-50\) $$\cos ^{2} \frac{5 \pi}{12}$$

Short Answer

Expert verified
\( \frac{2 - \sqrt{3}}{4} \)

Step by step solution

01

Identify Trigonometric Formulas

For the given problem, we can use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) to simplify the expression.
02

Set the Angle for Double Angle Formula

Use the angle \( \theta = \frac{5\pi}{12} \), then calculate \( 2\theta = \frac{10\pi}{12} = \frac{5\pi}{6} \).
03

Calculate\( \cos(\frac{5\pi}{6}) \)

The reference angle for \( \frac{5\pi}{6} \) is \( \frac{\pi}{6} \), where \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). Since \( \frac{5\pi}{6} \) is in the second quadrant where cosine is negative, \( \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \).
04

Substitute and Simplify

Substitute \( \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} \) into the formula: \[ \cos^2 \left( \frac{5\pi}{12} \right) = \frac{1 + (-\frac{\sqrt{3}}{2})}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} \text{ which simplifies to } \frac{2 - \sqrt{3}}{4}. \]

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Key Concepts

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

Double Angle Formula
The double angle formula is a powerful tool in trigonometry that helps us simplify expressions involving angles. In general, the double angle formulas are
  • For sine: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)
  • For cosine: \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta) \)
  • For tangent: \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \)
In practice, these formulas allow us to express trigonometric functions of double angles in terms of single angles. This simplifies complex problems, especially when dealing with squared trigonometric terms. For the exercise provided, we simplified \( \cos^2(\theta) \) by using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \). This is particularly useful for evaluating squares of cosine terms over specific intervals.
Trigonometric Functions
Trigonometric functions are fundamental concepts in trigonometry. They are based on the relationships between the angles and sides of right triangles. The main trigonometric functions are:
  • Sine (\(\sin\)): relates the opposite side to the hypotenuse in a right triangle.
  • Cosine (\(\cos\)): relates the adjacent side to the hypotenuse in a right triangle.
  • Tangent (\(\tan\)): relates the opposite side to the adjacent side in a right triangle.
Beyond right triangles, these functions extend to any angle, using the unit circle as a reference. Each function has specific properties such as periodicity, symmetry, and specific values at notable angles like \(\frac{\pi}{6}, \frac{\pi}{4},\) and \(\frac{\pi}{3}\). For instance, \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), which is crucial for solving our exercise as it helps in finding the cosine of angles in distinct quadrants.
Angle Simplification
Angle simplification is an essential skill when working with trigonometric identities. It involves reducing complex angles into simpler, well-known reference angles. The process often uses symmetries of trigonometric functions:
  • The unit circle helps visualize angles and their corresponding function values.
  • Reference angles are the acute angles that a terminal side makes with the horizontal axis of the unit circle.
  • Understanding quadrant rules always checks if the trigonometric functions are positive or negative.
In the provided problem, we simplified \(\frac{5\pi}{12}\) by using a double angle, \(2\theta = \frac{5\pi}{6}\). By recognizing that \(\frac{5\pi}{6}\) is in the second quadrant, we applied the property that cosine is negative here. Thus, finding \(\cos\left(\frac{5\pi}{6}\right)\) became straightforward with the reference angle \(\frac{\pi}{6}\). This understanding leads to accurate simplifications of trigonometric expressions.

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Most popular questions from this chapter

For a curve to be symmetric about the \(x\) -axis, the point \((x, y)\) must lie on the curve if and only if the point \((x,-y)\) lies on the curve. Explain why a curve that is symmetric about the \(x\) -axis is not the graph of a function, unless the function is \(y=0\) .

In Exercises \(7-10,\) write a formula for \(f \circ g \circ h\) $$ f(x)=\sqrt{x+1}, \quad g(x)=\frac{1}{x+4}, \quad h(x)=\frac{1}{x} $$

Evaluate \(\cos \frac{\pi}{12}\)

Match each equation with its graph. Do not use a graphing device, and give reasons for your answer. a. \(y=x^{4} \quad\) b. \(y=x^{7}\) \(\quad\) c. \(y=x^{10}\)

Graph the functions in Exercises \(37-56\) $$ y=\frac{1}{x}-2 $$
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