Problem 13
Question
In \(3-26,\) prove that each equation is an identity. $$ \frac{\cot \theta}{\csc \theta}=\cos \theta $$
Step-by-Step Solution
Verified Answer
The equation \( \frac{\cot \theta}{\csc \theta} = \cos \theta \) is an identity.
1Step 1: Recall Trigonometric Identities
Recall that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \). These will help us simplify the given equation: \( \frac{\cot \theta}{\csc \theta} = \cos \theta \).
2Step 2: Substitute Trigonometric Identities
Substitute the identities into the left-hand side of the equation: \( \frac{\cos \theta / \sin \theta}{1/ \sin \theta} \).
3Step 3: Simplify the Fraction
Simplify the fraction on the left-hand side by multiplying \( \frac{\cos \theta}{\sin \theta} \) by the reciprocal of \( \frac{1}{\sin \theta} \), which gives \( \cos \theta \times \sin \theta = \cos \theta \).
4Step 4: Compare Both Sides
Both sides of the equation simplify to \( \cos \theta \), which shows that \( \frac{\cot \theta}{\csc \theta} = \cos \theta \) is true for all \( \theta \) in the domain.
Key Concepts
Understanding CotangentExploring CosecantMastering Simplifying Trigonometric Expressions
Understanding Cotangent
The cotangent, often abbreviated as \( \cot \theta \), is one of the lesser-known trigonometric functions, but it plays a crucial role in many mathematical proofs and expressions. It is defined as the reciprocal of the tangent function. The tangent of an angle \( \theta \) in a right triangle is the ratio of the opposite side over the adjacent side, expressed as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Thus, the cotangent is:
- \( \cot \theta = \frac{1}{\tan \theta} \)
- \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
Exploring Cosecant
Cosecant, represented as \( \csc \theta \), is another reciprocal trigonometric function. It is the reciprocal of the sine function, meaning it is the ratio of the hypotenuse to the opposite side in the context of a right triangle. This can be mathematically expressed as:
- \( \csc \theta = \frac{1}{\sin \theta} \)
- \( \csc \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} \)
Mastering Simplifying Trigonometric Expressions
Simplifying trigonometric expressions involves rewriting them in a more manageable form, often by using basic trigonometric identities and reciprocal relationships. This process requires practice and a good understanding of the functions and their interactions. Here are a few tips:
- Use known identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) to transform expressions.
- Apply reciprocal identities, such as turning \( \frac{1}{\sin \theta} \) into \( \csc \theta \), to simplify complex fractions.
- Consider multiplying by the conjugate if you need to get rid of denominators in complex formulas.
Other exercises in this chapter
Problem 12
In \(3-17,\) find the exact value of \(\cos (A-B)\) for each given pair of values. \(A=45^{\circ}, B=270^{\circ}\)
View solution Problem 13
In \(3-17,\) find the exact value of \(\tan (A+B)\) and of \(\tan (A-B)\) for each given pair of values. $$ A=240^{\circ}, B=120^{\circ} $$
View solution Problem 13
\(\ln 3-17,\) find the exact value of \(\sin (A-B)\) and of \(\sin (A+B)\) for each given pair of values. \(A=30^{\circ}, B=270^{\circ}\)
View solution Problem 13
In \(3-17,\) find the exact value of \(\cos (A+B)\) for each given pair of values. \(A=60^{\circ}, B=270^{\circ}\)
View solution