sqrt(4 - (2sinx)^2)2cosx = 4cos^2(x)

There are steps in the middle (obviously) but I can't figure them out (Headbang)

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- Feb 28th 2009, 11:09 AMTYTYhow does this happen?
sqrt(4 - (2sinx)^2)2cosx = 4cos^2(x)

There are steps in the middle (obviously) but I can't figure them out (Headbang) - Feb 28th 2009, 11:51 AMReckoner
I can see why you are having trouble: the statement is

*not*an identity. For example, with $\displaystyle x=\pi,$ the left-hand side is $\displaystyle 2(-2)=-4,$ but the right-hand side is $\displaystyle 4(-1)^2=4.$ We can, however, show that the two expressions differ only in sign:

$\displaystyle 2\cos x\sqrt{4-(2\sin x)^2}$

$\displaystyle =2\cos x\sqrt{4-4\sin^2x}$

$\displaystyle =2\cos x\sqrt{4\left(1-\sin^2x\right)}$

$\displaystyle =4\cos x\sqrt{1-\sin^2x}$

$\displaystyle =4\cos x\sqrt{\cos^2x}$

$\displaystyle =4\cos x|\cos x|$

$\displaystyle =\pm4\cos^2x$ - Feb 28th 2009, 11:54 AMTYTY
This is very helpful and I appreciate it