# something's not working

• Feb 12th 2009, 03:51 PM
jahichuanna
something's not working
I have three problems, and none of them seem to work. they give us a hint to use trig identities.

48) dy/dx = sec^4 (x)

50) dy/dx = 4 cos^2 (x)

52) dy/dx = (cos^4 (x) - sin^4 (x))

I have no idea how to do any of these. I spent the last 50 minutes trying
• Feb 12th 2009, 03:59 PM
Jester
Quote:

Originally Posted by jahichuanna
I have three problems, and none of them seem to work. they give us a hint to use trig identities.

48) dy/dx = sec^4 (x)

50) dy/dx = 4 cos^2 (x)

52) dy/dx = (cos^4 (x) - sin^4 (x))

I have no idea how to do any of these. I spent the last 50 minutes trying

48) $\int \sec^4 x \,dx = \int \sec^2x \sec^2x \, dx$ let $u = \tan x$
50) $\cos^2x = \frac{1 + \cos 2x}{2}$

52) factor
$\cos^4 x - \sin^4 x = \left( \cos^2x - \sin^2x\right) \left( \cos^2x + \sin^2x\right) = \cos 2x$
• Feb 12th 2009, 04:00 PM
skeeter
$\sec^4{x} = \sec^2{x} \cdot \sec^2{x} = (1 + \tan^2{x})\sec^2{x}$

see u and du ?

$4\cos^2{x} = 4\left(\frac{1+\cos(2x)}{2}\right) = 2 + 2\cos(2x)$

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

• Feb 12th 2009, 04:25 PM
jahichuanna
$\cos^4{x} - \sin^4{x} = (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x} = \cos(2x)$

I don't see how you got that math.....

$(\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x}$
• Feb 12th 2009, 04:27 PM
skeeter
Quote:

Originally Posted by jahichuanna
$\cos^4{x} - \sin^4{x} = (\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x} = \cos(2x)$

I don't see how you got that math.....

$(\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^2{x} - \sin^2{x}$[/quote]

which part do you not "see"?
• Feb 12th 2009, 04:31 PM
jahichuanna
Quote:

Originally Posted by skeeter
I don't see how you got that math.....

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

wouldnt that work out to be

$(\cos^2{x} + \sin^2{x})(\cos^2{x} - \sin^2{x}) = \cos^4{x} - \sin^4{x}$?
• Feb 12th 2009, 04:35 PM
Slipery
Gotta remember that Cos2(x)+sin(2)x = 1
• Feb 12th 2009, 04:36 PM
jahichuanna
Quote:

Originally Posted by Slipery
Gotta remember that Cos2(x)+sin(2)x = 1

oh yeah.......ur right. i do need to study these.