Answer to Question #249349 in Trigonometry for Angela

Question

2n+1 > (n + 2) · sin(n) 

Answer 1

Let $P(n)$ be the proposition

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

$$2^{1+1}=4>3=1+2.$$

Inductive Step.

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

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

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

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

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

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

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

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

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

$$\sin(n)\leq1, n\in\R$$

Then for all positive integers $n$

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

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

Therefore we have proven that

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

for all positive integers $n.$