A. Is the answer
Hope this helped :)
Answer:
the answer is c
Step-by-step explanation:
The answer is (10,2),(14,-1), and (8,1)
Answer:
Unknown.
Step-by-step explanation:
Need more info.
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
