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jaejoon89
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I know the MacLaurin series for sin(x) but can't figure out how this reduces to the above (if it does).
How is sin(npi/2) = (-1)^((n-1)/2) for odd n?
- Thread starter jaejoon89
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SUMMARY
The equation sin(nπ/2) = (-1)^((n-1)/2) holds true for odd integers n. This relationship can be derived from the properties of the sine function and its periodicity. Specifically, sin(nπ/2) evaluates to -1 or 1 based on the value of n, confirming the validity of the formula. A proof by induction is a recommended method to establish this relationship rigorously.
PREREQUISITES- Understanding of trigonometric functions, particularly sine.
- Familiarity with the properties of periodic functions.
- Knowledge of mathematical induction as a proof technique.
- Basic comprehension of Maclaurin series expansions.
- Study the periodic properties of the sine function in detail.
- Learn about mathematical induction and its applications in proofs.
- Explore the Maclaurin series and its derivations for trigonometric functions.
- Investigate other trigonometric identities involving sine and cosine.
Mathematicians, educators, students studying trigonometry, and anyone interested in mathematical proofs and identities.
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