How Can the Maclaurin Series Validate a Trigonometric Inequality?

AI Thread Summary
The discussion revolves around proving the inequality 1 - t^2/2 ≤ cos(t) ≤ 1 for the interval 0 ≤ t ≤ 1. Participants express confusion about relating t to cos(t) and question whether the interval should be 0 ≤ t ≤ π/2 instead. The Maclaurin series for cos(t) is suggested as a potential method for validating the inequality. There is also a note that the problem may be more appropriate for a calculus forum rather than a precalculus one. The conversation highlights the need for clarity in problem statements and appropriate categorization of mathematical queries.
Clara Chung
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Homework Statement


show that 1-t^2/2 <=cos(t) <=1 for 0<=t<=1

Homework Equations


Trigonometry knowledge

The Attempt at a Solution


I don't know how to relate t with cos(t), and I also try to find out cos(1), but there is no result, so how can I start with this problem.
 
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Clara Chung said:

Homework Statement


show that 1-t^2/2 <=cos(t) <=1 for 0<=t<=1

Homework Equations


Trigonometry knowledge

The Attempt at a Solution


I don't know how to relate t with cos(t), and I also try to find out cos(1), but there is no result, so how can I start with this problem.
Are you sure that it's 0 ≤ t ≤ 1 , and not 0 ≤ t ≤ π/2 ?
 
Yes, this is what the question said. How can you solve for 0 <= t <= pi/2 ?
 
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Clara Chung said:
show that 1-t^2/2 <=cos(t) <=1 for 0<=t<=1
Have you know about Maclaurin series yet, and in particular, the Maclaurin series for cos(t)?

From another question you posted, you are taking a calculus class. If so, both questions should have been posted in the Calculus & Beyond section, not the Precalculus section.
 
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