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 sinθ and 2 cosθ 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 sinθ=sinθ+sinθ =∑_(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 cosθ=cosθ+cosθ =∑_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 sinθ and 2 cosθ 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 sinθ 2 cosθ=4sinθ cosθ . Therefore, the example required that, the concerned series be derived from each of the two components, that 2 sinθ 2 cosθ and 4 sinθ cos〖θ.〗 For convenience, the two expressions would be numbered as: (c) 2 sinθ 2 cosθ and (d) 4 sinθ cosθ. 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 sinθ 2 cosθ= ∑_(k=0)^θ▒(〖(-1)〗^2k θ^(4k+1))/(k+1/2)!K! (d) 4 sinθ sin〖θ=〗 ∑_(k=0)^θ▒(〖(-1)〗^2k θ^(4k+1))/(k+1/2)!K! ⇒ 2 sinθ 2 cosθ= 4sinθ 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 sinθ 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! ∴ 2sinθ 2 cosθ=∑_(k=0)^∞▒(〖(-1)〗^k θ^(4k+1))/(k+1/2)!k! Hence, (c) is verified. To verify (d): 4 sinθ 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! ∴ 4sinθ 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 sinθ 2 cosθ= 4 sinθ cos〖θ= ∑_(k=0)^θ▒((-1)^2k θ^(4k+1))/(k+1/2)!k! 〗 The above result mark the end of the article.

Article Source: http://www.abcarticledirectory.com

Yau Zakari www.silverbk.com’’>www.silverbk.com Silver Books Research scripts shop.

Still Searching? Last Chance to find what you're looking for with a Google Custom Search!

Or.... You can search this site using our Bing Custom Search!

Did You Like/Dislike This Article? Give It YOUR Rating!

Please Rate this Article

5 out of 54 out of 53 out of 52 out of 51 out of 5

No Ratings Yet. Be The First To Rate This Article

/EDF Publishing. All rights reserved. Script Services by: Sustainable Website Design Use of our free service is protected by our Privacy Policy and Terms of Service | Contact Us

Powered by ABC Article Directory