Prove that the average of the numbers n sin (n degrees) (n = 2, 4, 6, ... 180) is cot (1 degree)
Hint: Express everything in terms of sin and cos. Can you rearrange the result into a form that allows you to use some of the identities you've learned?
How to start? I know I have to use trigonometric identities but where?
Thanks a lot.
To prove that the average of the numbers n sin(n degrees) from n = 2 to n = 180 is cot(1 degree), we can start by expressing everything in terms of sin and cos.
First, let's consider the numbers n sin(n degrees). We can express this in terms of sin(n degrees) using the identity sin(n) = sin(n degrees):
n sin(n degrees) = n sin(n)
Next, let's express sin(n) in terms of sin(1 degree) using a trigonometric identity. We can use the identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B) with A = (n-1) degrees and B = 1 degree:
sin(n) = sin[(n-1) + 1] = sin(n-1)cos(1) + cos(n-1)sin(1)
Now, let's substitute this expression back into our original equation:
n sin(n degrees) = n [sin(n-1)cos(1) + cos(n-1)sin(1)]
Next, we expand the expression:
n sin(n degrees) = n sin(n-1)cos(1) + n cos(n-1)sin(1)
Now, we can rearrange this expression to a form that allows us to use identities. Let's express n-1 as (n-2+1):
n sin(n degrees) = n sin(n-2+1)cos(1) + n cos(n-2+1)sin(1)
Now, we can simplify this equation further. Using the identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B) with A = (n-2) degrees and B = 1 degree, we get:
n sin(n degrees) = n[sin(n-2)cos(1) + cos(n-2)sin(1)]cos(1) + n cos(n-2)sin(1)sin(1)
Continuing this process, we can expand and rearrange the expression until we reach the base case n=2.
Finally, we can prove that the average of n sin(n degrees) from n = 2 to n = 180 is cot(1 degree) by showing that the expression simplifies to cot(1 degree) at n=2.
I hope this helps you get started on the proof!
sin0+180sin180 = 0sin0 + 180sin0
2sin2 + 178sin178 = 2sin2 + 178sin2 = 180sin2
...
88sin88 + 102sin102 = 180sin88
so, we have
180(sin0 + sin2 + sin4 + ... + sin88) + 90sin90
= 180(sin0+sin2+...+sin90) - 90
That makes the average
2(sin0+sin2+...+sin90)-1
Gotta think some more. Back later. See what you can figure out.