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Circular Functions






     The importance and application of concepts of circular functions to solved real life problems need not be over emphasized. The reader would find below some useful research results about the properties of multiple sine and cosine function. The findings are simple but concrete results worth noticing. Although, these classes of functions are already well known but, the knowledge about them seems endless, as research advances, more facts are found and, the boundary of knowledge expands. In this present case, the series expression of: 2 sin⁡θ and 2 cos⁡θ was explored and presented below. The methodology of presentation is straight forward approach, this method is employed so as to avoid any ambiguity and, to give out more detailed explanation at the expense of rigor. There is no any special requirement needed for handling the script better than to recall the series expression for both sine and cosine function and, with little algebra. Finally, the paper may be of interest to scientist especially, those who engaged in the study and application of waves related issues and, educationist.
Now, much have been said about the introduction and, it will be best to start up the main content. So, there are two things to be shown as already hinted above and, are numbered: (a) and (b) respectively. Therefore, it would now be shown that:
(a): 2 sin⁡〖θ=∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(k+1/2)!〗
(b): 2 cos⁡〖θ=∑_(k=0)^∞▒(〖(-1)〗^k θ^2k)/2k!〗
Now, to verify the first one, that is (a);
Recall: 2 sin⁡θ=sin⁡θ+sin⁡θ
=∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(2k+1)! ∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(2k+1)!
=∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(2k+1)!+((-1)^k θ^(2k+1))/(2k+1)!
=∑_(k=0)^∞▒((-1)^k θ^(2k+1) + (-1)^k θ^(2k+1))/(2k+1)!
=∑_(k=0)^∞▒(2(-1)^k θ^(2k+1))/(2k+1)!
=∑_(k=0)^∞▒(2(-1)^k θ^(2k+1))/2(k+1/2)!
=∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(k+1/2)!
∴ 2sin⁡〖θ 〗=∑_(k=0)^∞▒((-1)^k θ^(2k+1))/(k+1/2)!
Hence, (a) is verified.
To verify (b), the same argument will be applied as that of (a) as follows:
Note that: 2 cos⁡θ=cos⁡θ+cos⁡θ
=∑_k^∞▒((-1)^k θ^2k)/2k!+∑_k^∞▒(〖(-1)〗^k θ^2k)/2k!
=∑_k^∞▒((-1)^k θ^2k)/2k!+(〖(-1)〗^k θ^2k)/2k!
=∑_k^∞▒((-1)^k θ^2k+ 〖(-1)〗^k θ^2k)/2k!
=∑_k^∞▒(2(-1)^k θ^2k)/2k!
=∑_k^∞▒((-1)^k θ^2k)/k!
∴ 2cos⁡〖θ =〗 ∑_k^∞▒((-1)^k θ^2k)/k!
Hence, (b) is verified.
Further more, some simple results regarding the product outcome for 2 sin⁡θ and 2 cos⁡θ is given below. The main purpose of this example is to show that, the product of these multiple functions have a unique series which described it or, represent it. And, the series can easily be worked out from any of the product expression component. Notice that the product expression can be represented as: 2 sin⁡θ 2 cos⁡θ=4sin⁡θ cos⁡θ . Therefore, the example required that, the concerned series be derived from each of the two components, that 2 sin⁡θ 2 cos⁡θ and 4 sin⁡θ cos⁡〖θ.〗 For convenience, the two expressions would be numbered as: (c) 2 sin⁡θ 2 cos⁡θ and (d) 4 sin⁡θ cos⁡θ. As mentioned earlier, the example is easy but worth mentioning because, some simple results can turn out to be obscure or difficult especially, when they are not first been shown to exist. Again, the absence of even a simple solution may have strong potential of blockage to certain progress, at lease for some time.
Without waste of time, it would now be shown that:
(c) 2 sin⁡θ 2 cos⁡θ= ∑_(k=0)^θ▒(〖(-1)〗^2k θ^(4k+1))/(k+1/2)!K!
(d) 4 sin⁡θ sin⁡〖θ=〗 ∑_(k=0)^θ▒(〖(-1)〗^2k θ^(4k+1))/(k+1/2)!K!
⇒ 2 sin⁡θ 2 cos⁡θ= 4sin⁡θ sin⁡〖θ=〗 ∑_(k=0)^θ▒(〖(-1)〗^2k θ^(4k+1))/(k+1/2)!K! .
To begin with, let (C) be taken for verification:
Recall: 2 sin⁡θ 2 cos⁡〖θ= ∑_(k=0)^∞▒(〖(-1)〗^k θ^(2k+1))/(k+1/2)!〗 .∑_(k=0)^∞▒(〖(-1)〗^k θ^2k)/k!
= ∑_(k=0)^∞▒(〖(-1)〗^k θ^(2k+1))/(k+1/2)! . (〖(-1)〗^k θ^2k)/k!
= ∑_(k=0)^∞▒(〖(-1)〗^k θ^(4k+1))/(k+1/2)!k!
∴ 2sin⁡θ 2 cos⁡θ=∑_(k=0)^∞▒(〖(-1)〗^k θ^(4k+1))/(k+1/2)!k!
Hence, (c) is verified.
To verify (d):
4 sin⁡θ cos⁡〖θ= 4[∑_(k=0)^θ▒((-1)^k θ^(2k+1))/(2k+1)! . ∑_(k=0)^θ▒((-1)^k θ^2k)/2k!]〗
= 4[∑_(k=0)^θ▒((-1)^k θ^(2k+1))/2(k+1/2)! . ((-1)^k θ^2k)/2k! ]
= 4[∑_(k=0)^θ▒((-1)^2k θ^(4k+1))/4(k+1/2)!k! ]
= ∑_(k=0)^θ▒(〖4(-1)〗^2k θ^(4k+1))/4(k+1/2)!k!
= ∑_(k=0)^θ▒((-1)^2k θ^(4k+1))/(k+1/2)!k!
∴ 4sin⁡θ cos⁡〖θ= ∑_(k=0)^θ▒((-1)^2k θ^(4k+1))/(k+1/2)!k! 〗
Hence, (d) is verified.
From results (c) and (d), it clearly indicated that:
2 sin⁡θ 2 cos⁡θ= 4 sin⁡θ cos⁡〖θ= ∑_(k=0)^θ▒((-1)^2k θ^(4k+1))/(k+1/2)!k! 〗
The above result mark the end of the article.






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Posted on 2014-01-25, By: *

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