Trignometric functions and their polynomial forms

Question

The trigonometric values also has a infinite series which I had learnt from the topics of Limits

Sin x = x - x³/ 3! + x⁵/ 5!.......

There are formulas like these for other trigonometric functions too. But here's where I don't understand. If we substitute 'π' instead of 'x' , then by using this formula , we get sin π not zero when it is zero. Similarly for all sin values of nπ ,where n ∈ Z get into the same trap.

sinπ = π - π³/3! + π⁵/5!...

0 = π - π³/3! + π⁵/5! ...

L.H.S ≠ R.H.S

This shouldn't happen, right ?

Where did I go wrong, then ?

Oh, thank you for clearing that. Can you explain about how sinπ is zero by using the series. Is it by limiting method ? — Parvathy, Jan 27 '21 at 05:18
Your last equation is wrong. Please read a little bit more about limits, sequences and series. There's no "polynomial form" for trigonometric functions. The three dots are very important, you can't forget about them. Learn what they mean — Yuriy S, Jan 27 '21 at 07:22

score 2 · Answer 1 · answered Jan 27 '21 at 07:13

2

Compute the partial sums $$S_p=\sum_{n=0}^p (-1)^n \frac{\pi^{2n+1}}{(2n+1)!}$$ For $p=4$, $S_4=0.00692527$, for $p=5$, $S_5=-0.00044516$, for $p=6$, $S_6=0.0000211426$.

answered Jan 27 '21 at 07:13

Claude Leibovici

260,315

So the value of sin π tends to zero. Is that what you mean, @Claude Leibovici ? – Parvathy Jan 27 '21 at 13:19

Trignometric functions and their polynomial forms

1 Answers1