Expert: Sherman D. - 11/30/2005

Question
Prove that the following:

sin2A/sinA – cos2A/sinA - cos2A/cosA = secA
and
cos4A = 8cos^4A - 8cos^2A + 1

Note: Please solve it in detail thanks.

if you know any informable website about the mathemmatics or the following above sbject please tell me.


Answer
By your first problem, did you mean

((sin(2A))/sinA) - ((cos(2A))/sinA) - ((cos(2A))/cosA) = secA

because when i try working it and graphing it, it doesn't come out right.

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cos(4A) = 8cosA^4 - 8cosA^2 + 1
cos(4A) = 2cos(2A)^2 - 1
cos(2A) = 2cos^2(A) - 1

cos(4A) = 2(2cosA^2 - 1)(2cosA^2 - 1) - 1
cos(4A) = 2(4cosA^4 - 2cosA^2 - 2cosA^2 + 1) - 1
cos(4A) = 2(4cosA^4 - 4cosA^2 + 1) - 1
cos(4A) = 8cosA^4 - 8cosA^2 + 2 - 1
cos(4A) = 8cosA^4 - 8cosA^2 + 2

info on this is found at www.mathpuzzle.com/Chebychev.html

for a shortcut, click ctrl+F and just type in cos(4A), and click next like twice.

Also go to
www.gomath.com/Questions/question.php?question=16081 and look under 16082, and look at the 5th line