I dont understand why sin(1+2n)(pi/2)=(-1)^n and sin(2n-1)(pi/2)=(-1)^n+1. Can you please explain it to me. Thank you
I dont understand why sin(1+2n)(pi/2)=(-1)^n and sin(2n-1)(pi/2)=(-1)^n+1. Can you please explain it to me. Thank you
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I dont understand why sin(1+2n)(pi/2)=(-1)^n and sin(2n-1)(pi/2)=(-1)^n+1. Can you please explain it to me. Thank you
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b In Problems 11-24 expand the gi x
A bartleby.com/solution-answer/chapter-113-problem-24e-differential-equations-with-boundary-value-problems-mindtap-course-list-9th-edition/9781337604918/in-problems-1124-expand-the-given-function-in-an-appropriate-cosine-or-sine-s. ☆
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< Chapter 11.3, Problem 24E >
Substitute the value of f (x) as cos x in the above equation.
2
an =
cos x cos
-x dx
*/2
cos x cos 2nx dx
пл
2
cos x cos
-x dx
*/2
Further solve the above equation.
a, = 4
cos (1+ 2n) x dx +
cos (2n – 1)x dr
sin (1+2n)x | %
(1+2n)
sin (2n-1)x
+
lo
(2n-1)
sin ( (1+2n)(들))
+
(1+2n)
sin ((2n-1)())
(2n-1)
For any non-negative value of n the value of the term sin ( (1 + 2n) ()) is
always 1 and the value of the term sin (1 – 2n) (5)) is 1 for even values
of n and –1 for odd values of n.
Further solve the above equation.
ao
(-1)"
(1+2n)
(-1)*+1
(2n-1)
Privacy - Terms
(n ie oven)
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