Question

Trig challenge!

Answer 1

Prove that the average of the numbers\[2\sin2^{\circ}, 4\sin 4^\circ , 6 \sin 6^{\circ}, \cdots, 180 \sin 180^{\circ}\]is \(90 \cot 1^{\circ}\).

Answer 2

hmm

Answer 3

This is just a little hard LOL

Answer 4

Hint: \[\sin(x)=\sin(180-x)\]

Answer 5

Just curious, when taking the average should we include the term 0*sin(0) to make the total number of terms 90 instead of 89?

Answer 6

That'd return the wrong average, but go ahead if you think you're onto something here.

Answer 7

Nahhh I just thought it looked nice haha.

Answer 8

But don't we already have 90 terms if you think about it?

Answer 9

Yeah you're right ahahaha. I'm just procrastinating from figuring out how to type my answer up and get the details right.

Answer 10

http://www.wolframalpha.com/input/?i=solve+y%3D%5Cint%5E180_2++x%28sind+x%29%5C%2Cdx%2C+for+y

Answer 11

http://www.wolframalpha.com/input/?i=10313%2F180%3D

Answer 12

wait you just took the average value of the function

Answer 13

and the interval length isn't even 180 so yeah, lol

Answer 14

i just tried something weird

Answer 15

Here I'll type up something real weird in a second, check this out.

Answer 16

http://www.wolframalpha.com/input/?i=%28cotd+1%29%3D

Answer 17

try to establish an equality and work with both sides

Answer 18

For maximum impressiveness make this substitution\[\Large \Re \{ \frac{\partial}{\partial t}\left(-2e^{i nt}\right)\}|_{t=\frac{\pi}{90}}=2n \sin(2n)\] into this sum \[\Large \frac{1}{90} \sum_{n=1}^{90}2n \sin(2n)\] push the summation through everything to get: \[\Large \frac{-2}{90}\Re \{ \frac{\partial}{\partial t}\left(\sum_{n=1}^{90}e^{i nt}\right)\}|_{t=\frac{\pi}{90}}\] The inner most term is a geometric series. The rest should be pretty self explanitory I think. =P

Answer 19

wha

Answer 20

@Kainui I am doing exactly that.

Answer 21

\[\begin{align}\sum\limits_{n=1}^{90} 2n\sin(2n) &=
90\sin 90+\sum\limits_{n=1}^{44} 180\sin(2n) \\~\\
&=90+\dfrac{90}{\sin 1}\sum\limits_{n=1}^{44} 2\sin(2n)\sin 1 \\~\\
&=90+\dfrac{90}{\sin 1}\sum\limits_{n=1}^{44} \cos(2n-1)-\cos(2n+1) ~~\color{red}{\star}\\~\\
&=90+\dfrac{90}{\sin 1}[\cos 1 - \cos 89] \\~\\
&=90\left(1+\dfrac{\cos 1 - \cos 89}{\sin 1}\right) \\~\\
& = 90\cot 1
\end{align}\]

\(\color{Red}{\star}\) : telescoping sum
\(\sum\limits_{r=1}^n f(r) - f(r+1) = f(1) - f(n+1) \)

Answer 22

good thing i didnt try .-.