# how does this happen?

• February 28th 2009, 12:09 PM
TYTY
how 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)
• February 28th 2009, 12:51 PM
Reckoner
Quote:

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 (Headbang)

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:

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

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

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

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

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

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

$=\pm4\cos^2x$
• February 28th 2009, 12:54 PM
TYTY
This is very helpful and I appreciate it