Prove that 2^n+1>(n+2).sin(n)​

Step-by-step explanation:

F(n)=|sin(n)|+|sin(n+1)|

then

F(n+π)=|sin(n+π)|+|sin(n+π+1)|=|sin(n)|+|sin(n+1)|=F(n)

and

F(π−n)=|sin(π−n)|+|sin(π−n+1)|=|sinn|+|sin(n−1)|≠F(n)

so we must prove when n∈(0,π), have

F(n)>2sin12

when n∈(0,π−1),then

F(n)=sinn+sin(n+1)=sinn(1+cos1)+sin1cosn

and n∈(π−1,π),then

F(n)=sinn−sin(n+1)

How prove it this two case have F(n)>2sin12? Thank you

and I know this well know inequality

|sinx|+|sin(x+1)|+|sin(x−1)|≥2sin1,x∈R

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

umm im not really sure because i never had to do this

step-by-step explanation: