# how does this happen?

Printable View

• Feb 28th 2009, 11:09 AM
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)
• Feb 28th 2009, 11:51 AM
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 $\displaystyle x=\pi,$ 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 AM
TYTY
This is very helpful and I appreciate it