#a_n = sin(pi n/6)+cos(pi n)# ?

Question

Charles Anctil · Accepted Answer

See below.

#a_n = sin(pi n/6)+cos(pi n)#

This is a periodic sequence in which

#a_n = a_(n+12)# with minimum and maximum values such that

#-2 le a_n le 1+sqrt(3)/2#

Attached a sequence plot

Scarlett Allison · Answer

undefined To find the general term $ a_n $ for the given sequence, we have:

$$ a_n = \sin\left(\frac{\pi n}{6}\right) + \cos(\pi n) $$

The terms involve trigonometric functions, and the values of $ \sin $ and $ \cos $ are periodic with period $ 2\pi $.

For $ \sin\left(\frac{\pi n}{6}\right) $:
The period is $ 2\pi $, but in this case, the period becomes $ 12\pi $ due to the coefficient $ \frac{\pi}{6} $. Thus, for integer values of $ n $, $ \sin\left(\frac{\pi n}{6}\right) $ will repeat every 12 terms.

For $ \cos(\pi n) $:
The cosine function has a period of $ 2\pi $. Thus, for integer values of $ n $, $ \cos(\pi n) $ will repeat every 2 terms.

To find the general term $ a_n $, we can determine the values for a few terms and observe the pattern:

For $ n = 0 $:
$$ a_0 = \sin(0) + \cos(0) = 0 + 1 = 1 $$

For $ n = 1 $:
$$ a_1 = \sin\left(\frac{\pi}{6}\right) + \cos(\pi) = \frac{1}{2} - 1 = -\frac{1}{2} $$

For $ n = 2 $:
$$ a_2 = \sin\left(\frac{\pi}{3}\right) + \cos(2\pi) = \frac{\sqrt{3}}{2} + 1 = \frac{\sqrt{3} + 2}{2} $$

By observing the pattern, we can generalize the terms as follows:

For $ n $ even:
$$ a_n = \sin\left(\frac{\pi n}{6}\right) + \cos(\pi n) $$

For $ n $ odd:
$$ a_n = \sin\left(\frac{\pi n}{6}\right) + \cos(\pi n) $$

Thus, the general term for the sequence $ a_n $ is:
$$ a_n = \sin\left(\frac{\pi n}{6}\right) + \cos(\pi n) $$