Chemical Engineering.

Lesson 0 - High School Stuff (Part 7: Trigonometric Identities)
A trigonometric identity is a relation betweem two different mathematical functions that involve trigonometric functions such as sin(x). They come in handy when simplifying mathematical expressions.

So you know how sin(x)=opposite/hypotenuse of an angle on a right-angle triangle, right? And cos(x)=adjacent/hypotenuse, and tan(x)=opposite/adjacent?


Well there's three others:

• csc(x)=H/O=1/sin(x). This is the cosecant.
  
• sec(x)=H/A=1/cos(x). This is the secant.
  
• cot(x)=A/O. This is the cotangent.
  

Hey, let's look back to Pythagorus's Theorem. O^2+A^2=H^2. We can get three trigonometric identities from this:

• Divide by H^2 on both sides. That gets you [sin(x)]^2+
  [cos(x)]^2=1.
  
• Divide by A^2 on both sides. That gets you [tan(x)]^2+1=[sec(x)]^2.
  
• Divide by O^2 on both sides. That gets you 1+[cot(x)]^2=[csc(x)]^2.
  

Honestly, only the first of those really gets used.

Another useful set of identities is the dual-angle identities. When you have a trigonometric function where the input involves two angles, you can convert it into a different expression, and vice versa.



These identities are helpful when, for example, you know the value of one angle but not the other.

And I THINK that covers just about all the high school stuff needed to understand Chem. Eng.

I think I'll start with the university-level math next, and then some basic kinematics.

why not just edit the other post?

Here we go

Applied Mathematics - Integration
So sometimes you wanna calculate the area under the curve. Specifically, a certain portion of the area under a curve, stretching from one x-value to another. An integral. This can be reinterpreted as calculating the area of an infinite number of infinitely thin rectangles that each extend to the curve.

With each rectangle, you're multiplying a y-value by an infinitely small difference in two x-values. Sound familar? It's essentially the opposite of a derivative, where you're dividing by a small difference in x-values. This is why calculating an integral means finding the antiderivative of the equation you're given.

Example: x^n, where n is a constant that isn't -1. The derivative would be n*x^(n-1). So the antiderivative would be [x/(n+1)]*x^(n+1).

• If the integral you're calculating has two specific end points, you calculate the value ls of the integral equation at both end points, and then you subtract the value at the first end point from the value at the second end point.
  
• If you have no specific end points, you need to add a nondescript constant to the integral equation, because something like that won't affect the original derivative.

calculus is so sick. math is just a beaut dude.