Need help verifying identities

**laoatt** · October 22nd 2008, 05:35 PM

I was given a homework sheet to verify several identities, but I could not figure out these two problems. Can someone enlighten me.

1. cos 4x= 1-8sin²x cos²x

2. (sin³x - cos³x)/(sinx - cosx) = 1+sinxcosx

**Shyam** · October 22nd 2008, 05:51 PM

1. cos 4x= 1-8sin²x cos²x

Right Side = 1 - 2 (4sin²x cos²x) = 1 - 2 (2sin x . cos x)²

= 1 - 2 (sin 2x)² = (1)- 2 sin² 2x = (sin² 2x + cos² 2x) -2sin² 2x

= cos² 2x - sin² 2x = cos 4x

2. (sin³x - cos³x)/(sinx - cosx) = 1+sinxcosx

Left side $= \frac {\sin^3 x -\cos^3 x}{\sin x - \cos x}$

$= \frac {(\sin x -\cos x)(\sin^2 x +\sin x . \cos x + \cos^2 x)}{\sin x - \cos x}$

because $a^3-b^3 = (a-b)(a^2+ab+b^2)$

$= (\sin^2 x + \cos^2 x + \sin x . \cos x )$

$= 1+ \sin x . \cos x$

**skeeter** · October 22nd 2008, 05:54 PM

$\cos(4x) =$

$\cos[2(2x)] =$

$1 - 2\sin^2(2x) =$

$1 - 2(2\sin{x}\cos{x})^2 =$

finish it.

hint for the second one ...

$a^3 - b^3 = (a-b)(a^2+ab+b^2)$

**laoatt** · October 22nd 2008, 07:34 PM

ahh, I see now. Thank you guys.

**toraj58** · October 24th 2008, 02:39 PM

from left:

$\cos4x = \cos2(2x)$
$= \cos^22x - \sin^22x$
$= (\cos^2x-sin^2x)^2 - (2\sin x\cos x)^2$
$= (\cos^4x+\sin^4x)-2\cos^2x\sin^2x-4\sin^2x\cos^2x$

since we have: $(\cos^4x+\sin^4x) = 1-2\sin^2x\cos^2x$
but why:
simply we can approve it like this:

$(\sin^2x+\cos^2x)^2 = \sin^4x+\cos^4x+2\sin^2x\cos^2x$
and we know that: $(\sin^2x+\cos^2x)$ = 1

we replace it and:

$= (1-2\sin^2x\cos^2x)-2\cos^2x\sin^2x-4\sin^2x\cos^2x$

finally we simplifying it:

$= 1-8\sin^2x\cos^2x$

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