how does this happen?

**TYTY** · February 28th 2009, 11:09 AM

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

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

**Reckoner** · February 28th 2009, 11:51 AM

Originally Posted by TYTY

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

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

I can see why you are having trouble: the statement is not an identity. For example, with $x=\pi,$ the left-hand side is $2(-2)=-4,$ but the right-hand side is $4(-1)^2=4.$ We can, however, show that the two expressions differ only in sign:

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

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

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

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

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

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

$=\pm4\cos^2x$

**TYTY** · February 28th 2009, 11:54 AM

This is very helpful and I appreciate it

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