/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Evaluate each expression exactly... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each expression exactly, if possible. If not possible, state why. $$\csc ^{-1}\left[\csc \left(\frac{15 \pi}{4}\right)\right]$$

Short Answer

Expert verified
The expression evaluates to \(-\frac{\pi}{4}\).

Step by step solution

01

Understand the Problem

We need to evaluate the expression \( \csc^{-1}\left[\csc\left(\frac{15 \pi}{4}\right)\right] \). The goal is to find the exact value or determine if it's not possible.
02

Simplify the Argument of \(\csc\)

Simplify \( \frac{15 \pi}{4} \) by expressing it in terms of a coterminal angle within the interval \( [0, 2\pi] \).\[ \frac{15\pi}{4} = 2\pi + \frac{7\pi}{4} \]Thus, \( \frac{15\pi}{4} \) is coterminal with \( \frac{7\pi}{4} \).
03

Evaluate the Cosecant Function

Find \( \csc\left(\frac{7\pi}{4}\right) \).Since \( \frac{7\pi}{4} = 2\pi - \frac{\pi}{4} \), it is in the fourth quadrant where the sine is negative and \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \). Thus, \( \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).Therefore, \( \csc\left(\frac{7\pi}{4}\right) = \frac{-1}{\sin\left(\frac{7\pi}{4}\right)} = -\sqrt{2} \).
04

Apply Inverse Cosecant

Now, evaluate \( \csc^{-1}(-\sqrt{2}) \).Since \( \csc^{-1}x \) is defined for \( x \leq -1 \) or \( x \geq 1 \), and the range of \( \csc^{-1} \) is \( [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] \), we need an angle whose cosecant is \(-\sqrt{2}\). The commonly known angle is \( -\frac{\pi}{4} \), as it is in the fourth quadrant and returns \( -\sqrt{2} \) for cosecant.

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Key Concepts

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

Cosecant Function
The cosecant function, denoted as \( \csc \), is the reciprocal of the sine function. This means that for any angle \( \theta \), the equation \( \csc(\theta) = \frac{1}{\sin(\theta)} \) holds true.
Understanding this reciprocal relationship helps in solving trigonometric equations where sine and cosecant are involved.Key points about the cosecant function:
  • It is undefined for angles where sine equals zero. This is because you cannot divide by zero.
  • The range of the cosecant function is expressed as \( (-\infty, -1] \cup [1, \infty) \), which means it does not take any values between -1 and 1.
  • This function has a periodic cycle of \( 2\pi \), similar to sine, meaning it repeats its values every \( 2\pi \) radians.
When working with cosecant, being able to determine the sine of an angle is crucial as it directly determines the value of its reciprocal function. In the context of the exercise, by finding \( \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2} \), we could easily compute \( \csc(\frac{7\pi}{4}) = -\sqrt{2} \).
Co-terminal Angles
Co-terminal angles are angles that share the same terminal side, essentially differing from each other by full circles, which corresponds to \( 2\pi \) radians or \( 360\) degrees. This means that two angles are co-terminal if their standard angle difference is a multiple of \( 2\pi \).
The main method to find a co-terminal angle is to either add or subtract \( 2\pi \) to the given angle until you reach an angle that falls within the desired interval, usually \([0, 2\pi)\).
  • For example, to find a co-terminal angle for \( \frac{15\pi}{4} \), you can subtract \( 2\pi \) to get \( \frac{7\pi}{4} \).
  • This helps simplify trigonometric expressions as it allows the angle to be viewed in its simplest form.
In the exercise at hand, the angle \( \frac{15\pi}{4} \) was reduced to \( \frac{7\pi}{4} \) to make the calculations manageable. By using co-terminal angles, complex trigonometric evaluations become simpler, effectively reducing any elongated circle measurements.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables involved where the functions are defined. These identities are immensely useful for simplifying trigonometric expressions and solving equations.
Some commonly used trigonometric identities include:
  • The Pythagorean identities, such as \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
  • The reciprocal identities: \( \csc(\theta) = \frac{1}{\sin(\theta)} \) and \( \sec(\theta) = \frac{1}{\cos(\theta)} \).
  • Co-function identities, like \( \sin(\frac{\pi}{2} - \theta) = \cos(\theta) \).
In the context of the exercise, the reciprocal identity was key. By understanding \( \csc(\theta) = \frac{1}{\sin(\theta)} \), one could directly find the inverse cosecant through simplification processes. Knowing these identities allows for more efficient problem-solving and can lead to elegant solutions by substituting equivalent expressions.

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Most popular questions from this chapter

In calculus, the method of separation of variables is used to solve certain differential equations. Given an equation with two variables, the method consists of writing the equation in such a way that each side of the equation contains only one type of variable. Use the product-to-sum and sum-to-product identities to separate the variables \(x\) and \(y\) in each equation. $$2+\cos (x+y)=\cos (x-y)$$

Find the smallest positive values of \(x\) that make the statement true. Give the answer in degrees and round to two decimal places. $$\ln x-\sin x=0$$

Refer to the following: The figure below shows the graph of \(y=2 \cos x-\cos (2 x)\) between \(-2 \pi\) and \(2 \pi .\) The maximum and minimum values of the curve occur at the turning points and are found in the solutions of the equation \(-2 \sin x+2 \sin (2 x)=0\). CAN'T COPY THE GRAPH Solve for the coordinates of the turning points of the curve between 0 and \(2 \pi\)

Solve each trigonometric equation on \(0^{\circ} \leq \theta<360^{\circ} .\) Express solutions in degrees and round to two decimal places. $$\frac{1}{16} \csc ^{2}\left(\frac{x}{4}\right)-\cos ^{2}\left(\frac{x}{4}\right)=0$$

Explain the mistake that is made. $$\text { Solve } \sqrt{1+\sin x}=\cos x \text { on } 0 \leq x \leq 2 \pi$$ Solution: Square both sides. \(\quad 1+\sin x=\cos ^{2} x\). Use the Pythagorean identity. \(\quad 1+\sin x=\underbrace{\cos ^{2} x}_{1-\sin ^{2} x}\). Simplify. \(\sin ^{2} x+\sin x=0\). Factor. \(\quad \sin x(\sin x+1)=0\). Set each factor equal to zero.\(\sin x=0 \quad\) or \(\quad \sin x+1=0\). Solve for \(\sin x, \quad \sin x=0 \quad\) or \(\quad \sin x=-1\). Solve for \(x\) \(x=0, \pi, \frac{3 \pi}{2}, 2 \pi\). This is incorrect. What mistake was made?

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