Prove that $2^{n+1}>(n+2) \cdot \sin(n)$ for all positive integers n.

Question

Proof: If $P(n)$ represents the given proposition.

(1) Basic Step: $P(n)$ for $n=1$ is $$2^{1+1}>(1+2) \cdot \sin(1)$$ $$2^{2}>3 \cdot \sin(1)$$ $$4>3 \cdot \sin(1)$$ Since $\sin(n)≤1$ and $1<4/3$ and so $\sin(n)<4/3$ by putting $n=1$, we obtain $$\sin(1)<4/3 \implies 4>3 \cdot \sin(1)$$ Which is true.

(2) Induction Step: Suppose that $P(n)$ is true for $n=k$, i.e. Let $$2^{k+1}>(k+2) \cdot \sin(k)$$ To prove that $P(n)$ to be true for $n=k+1$.

$$2^{(k+1)+1}=2^{k+1}2=2\cdot2^{k+1}>2(k+2) \cdot \sin(k) $$

How to prove $P(n)$ to be true for $n=k+1$. I got stuck here. Would appreciate for your assistance. Also review my proof if there is any mistake while writing.

Answer 1

Here classical induction needs some modification.

Step 1. Show that it is true for $n=1$.

Step 2. Since $n+2\ge (n+2)\sin n$, it suffices to show that $$ 2^{n+1}>n+2 $$ for $n\ge 2$, which can be inductively.