View Full Version : sadistic calculus professor

We're doing volumes of rotated functions. I've got most of them done, but I just have to solve for x in this equation, and NO identities are coming to mind... and I can't find any that apply in my lists or on the internet. Can anybody see a way to solve for x here? (Algebraically, not graphically...)

9 cos(x)

-- = -------

16 sin(2x)

tabstop

09-06-2008, 08:58 AM

NO identities are coming to mind? You can't find an identity for sin(2x)?

Wow I'm stupid... It's been 3 years since my last math class.

indigo0086

09-06-2008, 05:49 PM

dude I hear you. I took prob and stat last semester after roughly 2 or three years of not taking any math courses and felt like an idiot because I couldn't do integrals or basic stuff. I remembered quickly with the power of google and the like.

Kernel Sanders

09-06-2008, 06:36 PM

sin(2x) = 2*sin(x)*cos(x)

9/16 = cos(x) / (2 * sin(x) * cos(x))

9/8 = 1/sin(x)

sin(x) = 8/9

x = arcsin(8/9)

BobMcGee123

09-07-2008, 07:21 PM

Wow I'm stupid... It's been 3 years since my last math class.

Don't worry about it man, once you get back into the swing of things *pushes up glasses...pockets calculator* this math stuff will be nothing but a reflex for ya.

twomers

09-08-2008, 04:59 AM

It's worse when they give you something like:

x + sin( x ) = 0.3

or something. Can't solve it analytically and it drives me mad!

OnionKnight

09-12-2008, 05:59 AM

When you have to integrate sin(x)^8 dx is when I go SCREW YOU

abachler

09-12-2008, 10:56 AM

We're doing volumes of rotated functions. I've got most of them done, but I just have to solve for x in this equation, and NO identities are coming to mind... and I can't find any that apply in my lists or on the internet. Can anybody see a way to solve for x here? (Algebraically, not graphically...)

9 cos(x)

-- = -------

16 sin(2x)

-X / 2 dX

they should have covered the relation between sine and coine as derivatives of each other.

Yes, sine and cosine have a relationship as derivatives, but when one is a function of x and the other is a function of 2x, it's a more complex relationship. The way to simplify it was to use sin(2x) = 2sin(x)cos(x), and therefore the whole expression simplifies to 1/2sin(x).

Powered by vBulletin® Version 4.2.5 Copyright © 2019 vBulletin Solutions Inc. All rights reserved.