LoyalKnight
New member
- Joined
- Nov 14, 2013
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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?
So far I've gotten:
We first find the sum,
so 2 sin 2 + ..... 180 sin 180
we add 0 sin 0 to teh sum because 0 sin 0 = 0
Then we get 0 sin 0 + .... 180 sin 180
sin(180-x) = sin(x)
so we get 0 sin 0 + 180 sin 180 = 0 sin 0 + 180 sin 0
also we pair up all the other terms likewise: 2 sin 2 + 178 sin 178 = 2 sin 2 + 178 sin2
the only one not paired up si 90 sin 90
We factor a 180 from the paired up terms :
180(sin 0 + sin 2....sin 88) + 90 sin 90
180 (sin 90) - 90 = 90 sin 90 so we get
180 ( sin 0 + ..... sin 90) - 90
Divide everything by 90 to get teh average
so we get 2 (sin 0 + .... sin 0) - 1
What do I do next? I'm confused. How do I apply trigonometric identities? I think I need to telescope but how?
Thanks in advance for help, it is appreciated.
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?
So far I've gotten:
We first find the sum,
so 2 sin 2 + ..... 180 sin 180
we add 0 sin 0 to teh sum because 0 sin 0 = 0
Then we get 0 sin 0 + .... 180 sin 180
sin(180-x) = sin(x)
so we get 0 sin 0 + 180 sin 180 = 0 sin 0 + 180 sin 0
also we pair up all the other terms likewise: 2 sin 2 + 178 sin 178 = 2 sin 2 + 178 sin2
the only one not paired up si 90 sin 90
We factor a 180 from the paired up terms :
180(sin 0 + sin 2....sin 88) + 90 sin 90
180 (sin 90) - 90 = 90 sin 90 so we get
180 ( sin 0 + ..... sin 90) - 90
Divide everything by 90 to get teh average
so we get 2 (sin 0 + .... sin 0) - 1
What do I do next? I'm confused. How do I apply trigonometric identities? I think I need to telescope but how?
Thanks in advance for help, it is appreciated.
