Problem 42
Question
Express the first trigonometric function in terms of the second. $$ \csc \theta, \cot \theta $$
Step-by-Step Solution
Verified Answer
\(\csc \theta = \cot \theta\)
1Step 1: Expressing \(\sin \theta\) in terms of \(\cot \theta\)
Start by considering the definition of \(\cot \theta\). By definition, we have \(\cot \theta = \frac{\cos \theta}{\sin \theta}\). \nNow, let's rewrite this in terms of \(\sin \theta\): We can isolate \(\sin \theta\) in the equation above by taking the reciprocal of both sides, which gives us \(\sin \theta = \frac{1}{\cot \theta}\), since the reciprocal of \(\cot \theta\) is \(1/\cot \theta\).
2Step 2: Expressing \(\csc \theta\) in terms of \(\cot \theta\)
Next, recall that \(\csc \theta\) is simply the reciprocal of \(\sin \theta\). So, \(\csc \theta = \frac{1}{\sin \theta}\).\n Using the result we obtained in the previous step, we can express \(\csc \theta\) in terms of \(\cot \theta\). Simply replace \(\sin \theta\) in the above equation with our result from step 1, and we obtain \(\csc \theta = \frac{1}{(1/\cot \theta)}\).
3Step 3: Simplifying the expression
Now simplify the expression \(\csc \theta = \frac{1}{(1/\cot \theta)}\). By recalling the rule that dividing by a fraction is the same as multiplying by its reciprocal, this expression simplifies to \(\csc \theta = \cot \theta\).
Key Concepts
Cosecant FunctionCotangent FunctionReciprocal Identities
Cosecant Function
The cosecant function, often represented as \( \csc \theta \), is one of the six fundamental trigonometric functions. It is the reciprocal of the sine function. In simpler terms, it means that the cosecant of an angle is the inverse of the sine of that angle.
\[ \csc \theta = \frac{1}{\sin \theta} \]
This identity highlights that whenever the sine of an angle is zero, the cosecant becomes undefined, as you cannot divide by zero.
Understanding the role of \( \csc \theta \) is crucial in trigonometry as it is often used to simplify expressions or to solve equations where \( \sin \theta \) is present.
\[ \csc \theta = \frac{1}{\sin \theta} \]
This identity highlights that whenever the sine of an angle is zero, the cosecant becomes undefined, as you cannot divide by zero.
Understanding the role of \( \csc \theta \) is crucial in trigonometry as it is often used to simplify expressions or to solve equations where \( \sin \theta \) is present.
Cotangent Function
The cotangent function \( \cot \theta \) is another key trigonometric function. It is defined as the reciprocal of the tangent function. That means the cotangent of an angle is the ratio of the adjacent side to the opposite side in a right-angled triangle, but it can also be expressed using sine and cosine.
\[ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \]
This definition shows that \( \cot \theta \) is undefined when \( \sin \theta = 0 \), because division by zero is invalid. Understanding \( \cot \theta \) can help in expressing other trigonometric functions, such as to simplify expressions involving both sine and cosine.
\[ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \]
This definition shows that \( \cot \theta \) is undefined when \( \sin \theta = 0 \), because division by zero is invalid. Understanding \( \cot \theta \) can help in expressing other trigonometric functions, such as to simplify expressions involving both sine and cosine.
Reciprocal Identities
Reciprocal identities are a set of trigonometric relationships between the basic trigonometric functions and their reciprocals. These identities are helpful because they allow for the transformation or simplification of expressions.
Here's a quick reminder of some key reciprocal identities:
By understanding these identities, you can maneuver through problems with greater ease and efficiency, making them valuable tools in both calculus and geometry.
Here's a quick reminder of some key reciprocal identities:
- \( \csc \theta = \frac{1}{\sin \theta} \)
- \( \sec \theta = \frac{1}{\cos \theta} \)
- \( \cot \theta = \frac{1}{\tan \theta} \)
By understanding these identities, you can maneuver through problems with greater ease and efficiency, making them valuable tools in both calculus and geometry.
Other exercises in this chapter
Problem 42
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