Math Problem Statement
Solution
Let's analyze the trigonometric problems provided in the image:
1. Problem Analysis
The task is to find the exact values of the trigonometric expressions using their periodic properties.
a)
The secant function, , is periodic with a period of . To find , we can first reduce it within the range of to :
Thus, .
Now, we know that:
And:
So:
Therefore:
b)
The cotangent function, , is periodic with a period of . To find , we first reduce it within the range of to :
Thus:
Now, we know:
Therefore:
Summary of Answers:
a)
b)
Would you like to go into more details about how these solutions were derived, or do you have any other questions?
Related Questions:
- What is the period of the tangent and cotangent functions?
- How do you find the exact values of trigonometric functions at specific angles?
- What are the co-function identities for sine and cosine?
- How do you determine the signs of trigonometric functions in different quadrants?
- How are trigonometric functions related to the unit circle?
Tip:
When working with trigonometric functions, always remember to reduce angles using their periods to find equivalent values within a standard range (e.g., to or to ).
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Periodic Functions
Secant
Cotangent
Formulas
sec(x) = 1/cos(x)
cot(x) = 1/tan(x)
Period of sec(x) = 360°
Period of cot(x) = π
Theorems
Periodic Properties of Trigonometric Functions
Exact Values of Trigonometric Functions
Suitable Grade Level
Grades 10-12