Using the trigonometric identity of sin &#x2061;<!-- ⁡ --> 2 &#x03B1;<!-- α --> =

taghdh9

taghdh9

Answered question

2022-06-19

Using the trigonometric identity of sin 2 α = 2 sin α cos α, I rewrote the expression to:
lim n ( sin ( 2 1 ) n 1 cos 1 + + sin ( 2 n ) n n cos n ) = lim n ( 2 sin ( 1 ) n 1 + + 2 sin ( n ) n n )
I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:
2 sin ( 1 ) n 1 + + 2 sin ( n ) n n 2 n 1 + + 2 n n 2 n n 2
What am I missing?

Answer & Explanation

Brendon Fernandez

Brendon Fernandez

Beginner2022-06-20Added 14 answers

The inequality
k = 1 n 1 k 1 + 1 n 1 x d x 2 n ,
yields
| 2 sin ( 1 ) n 1 + . . . + 2 sin ( n ) n n | 4 1 n 0.
Note that sin ( ) can change signs, so absolute value is needed in the estimate.

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