as u could see,that sqrt.(6) is a non-recurring & non-terminating decimal ,so this problem arises.
Having a simple problem again.
First: I'm really sorry, I've looked and I can't find and post on how to insert the correct "Math-Code" (well, I don't know how to insert degrees, square root and other things).
* = degrees
Determine the EXACT value of cos(15*) by rewriting cos(15*) = cos(45*-30*):
I can get exact values like:
cos15* = cos(45*-30*) = cos45*cos30* + sin45*sin30* = 0.965925......
But, ofcourse, this won't do. The correct answer is:
(squareroot(6) + squareroot(2))/4
How can I get the correct (well, I guess both are correct if I write the whole answer, but I mean the "better looking" one) later answer instead of the first? Appreciate help!!
Hi, and thank you for your response.
I'm not sure if I understand you correct. You mean that sqrt.(6) is a never-ending decimal right?
I probably wasn't clear with my problem, or . I can get the correct answer, but I don't know how to calculate/convert the result so that it shows (sqrt(6) + sqrt(2))/4 instead of 0.965925......
I know how to calculate cos(45*-30*) = cos45*cos30*+sin45*sin30* = =0.965925......
This is how the problem is solved in the book:
cos(45* - 30*)= (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) =
= (sqrt(3) + 1)/(2sqrt(2)) = sqrt(2)(sqrt(3) + 1)/ = (sqrt(6)+sqrt(2))/4
My issue is that I don't know how to rewrite the cos45*cos30*+sin45*sin30* into (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2)
and so on.
I'm sorry if my explanation is bad. I haven't studied much math before.
Again, appreciate help!
Edit: From your latest post it seems that you do not know your unit circle, so I suggest you review that section in the book. Wikipedia has an overview. These values are found using special properties of and triangles, discussions of which can be found in topics 6 and 7 on this page.
The strange thing is that the book never mentioned (yet) that it's important to know these specific values, so I thought there was some almost-impossible magic way to get these numbers by doing some kind of calculation.
As said before, I'm a math-beginner