Problem 57
Question
Explain why there are no real numbers \(x\) satisfying the equation \(9 \csc x=1\).
Step-by-Step Solution
Verified Answer
There are no real solutions because the sine function's range is [-1, 1], which does not include 9.
1Step 1: Convert to Sine Function
The equation is given as \(9 \csc x = 1\). We know \(\csc x\) is the reciprocal of \(\sin x\), so \(\csc x = \frac{1}{\sin x}\). Substitute this in the equation: \(9 \times \frac{1}{\sin x} = 1\).
2Step 2: Simplify the Equation
Simplify the equation from Step 1: \(\frac{9}{\sin x} = 1\). To clear the fraction, multiply both sides by \(\sin x\) to get \(9 = \sin x\).
3Step 3: Analyze the Sine Function Values
The sine function \(\sin x\) has a range of values between -1 and 1, inclusive. Since \(9\) is outside of this range, \(\sin x = 9\) is not possible for a real number \(x\).
4Step 4: Conclusion
Since the value \(9\) is not in the range of the sine function, there are no real solutions \(x\) that satisfy the equation \(9 \csc x = 1\).
Key Concepts
Cosecant FunctionSine FunctionRange of Trigonometric Functions
Cosecant Function
The cosecant function, denoted as \( \csc x \), is one of the six primary trigonometric functions. It is defined as the reciprocal of the sine function. Therefore, \( \csc x = \frac{1}{\sin x} \). Recall that for all angles \( x \) where \( \sin x eq 0 \), the cosecant function is defined. This function is particularly useful because it helps in solving equations and understanding relationships in trigonometry.When dealing with reciprocal values like the cosecant, it’s important to remember that the cosecant function will not exist when the sine function equals zero. This is because dividing by zero is undefined. For example:
- If \( \sin x = 0 \), \( \csc x \) is undefined.
- Common angles where \( \csc x \) is defined are multiples of \( \frac{\pi}{2} \).
Sine Function
The sine function, represented as \( \sin x \), is a fundamental concept in trigonometry. This function is about finding the y-coordinate of a point on the unit circle corresponding to an angle \( x \).Sine functions are periodic and their values repeat every \( 2\pi \) radians. The sine curve oscillates smoothly between -1 and 1. This oscillation results from its definition based on circular movement.When we explore equations involving the sine function, we often need to remember:
- The sine of any angle is computed as: \( \sin x = \frac{\text{opposite}}{\text{hypotenuse}} \) in a right triangle.
- The range of the sine function is from -1 to 1.
- For instance, \( \sin 0 = 0 \) and \( \sin \frac{\pi}{2} = 1 \).
Range of Trigonometric Functions
The range of trigonometric functions refers to the set of possible output values these functions can take. This concept is key to understanding why certain equations, like \( 9 \csc x = 1 \), have no solutions.Focusing on the sine function, its range is limited to values between -1 and 1, inclusive. This means:
- Any real number input to the sine function will result in an output somewhere within this range.
- If an equation suggests a sine output outside this range, it signals no real \( x \) can satisfy it.
Other exercises in this chapter
Problem 56
Putting the Shot The range of a shot put released from a height \(h\) above the ground with an initial velocity \(v_{0}\) at an angle \(\theta\) to the horizont
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Verify the given identity. $$ \sec 2 x=\frac{1}{2 \cos ^{2} x-1} $$
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Verify the given identity. $$ \frac{\cos (-t)}{1+\tan (-t)}-\frac{\sin (-t)}{1+\cot (-t)}=\sin t+\cos t $$
View solution Problem 57
Find the first three \(x\) -intercepts of the graph of the given function on the positive \(x\) -axis. $$ f(x)=\sin x+\tan x $$
View solution