Answer to Question #249349 in Trigonometry for Angela

Question #249349

2n+1 > (n + 2) · sin(n) 


1
Expert's answer
2021-10-11T11:11:07-0400

Let P(n)P(n) be the proposition

Basis Step. P(1)P(1) is true, because

21+1=4>3=1+2.2^{1+1}=4>3=1+2.

Inductive Step.

Assume that P(k)P(k) holds for an arbitrary positive integer k.k. That is, we assume that


2k+1>k+22^{k+1}> k + 2

Under this assumption, it must be shown that P(k+1)P(k + 1) is true, namely, that


2(k+1)+1>(k+1)+22^{(k+1)+1}> (k+1) + 2

2(k+1)+1=2(2k+1)2^{(k+1)+1}=2(2^{k+1})

=2k+1+2k+1>k+2+k+2=2^{k+1}+2^{k+1}>k+2+k+2

=k+3+k+1>k+3=k+3+k+1>k+3

This shows that P(k+1)P(k + 1) is true under the assumption that P(k)P(k) is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for all positive integers n.n. That is, we have proven that 2n+1>n+22^{n+1}> n + 2 for all positive integers n.n.

sin(n)1,nR\sin(n)\leq1, n\in\R

Then for all positive integers nn

(n+2)sin(n)n+2,nR(n+2)\cdot\sin(n)\leq n+2, n\in\R

We have proven that 2n+1>n+22^{n+1}> n + 2 for all positive integers n.n.

Therefore we have proven that

2n+1>(n+2)sin(n)2^{n+1}>(n+2)\cdot\sin(n)

for all positive integers n.n.



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