Trig fun

[ResidentBiscuit]

Ok, so this is killing me. I'm taking this damn online trig class, and the professor expects us to have "mastered" algebraic manipulation of trig functions as a pre req to this course. No where in my education have I had to do this.

Anyway, I've been trying to figure it out but this just makes no sense. For example:

Somehow this is true

sec^2(x)/sec(x) = sec(x)

How? I have absolutely no idea. And to make things worse, nowhere are the rules of manipulating trig functions in this course, since we were supposed to "have it mastered by now."

If I just have a rule sheet, I'd be good to go. But nope, I'm sitting here trying to Isaac Newton my way through all this damn problems.

[atropos]

sec^2(x)/sec(x) = sec(x) 

sec^2(x) = sec(x) * sec(x)

sec^2(x) = sec^2(x)

QED

Try looking at it terms of functions. If the problem was stated :

Prove that [f(x)]2/ f(x) = f(x)

would you be able to solve it then?
Trig functions are just that - functions. (Albeit sometimes with an infinite number of discontinuities)

[ResidentBiscuit]

Oh wow, well that was a little too obvious. Guess I got frustrated enough I forgot basic math. I don't know why, but this crap is just throwing me off.

sec^4(x) - 2 sec^2(x) tan^2(x) + tan^4(x)

Supposed to factor this, all the answers are pretty small. Not sure how to get there considering there isn't one common factor in these terms (that I see, at least)

[atropos]

sec^4(x) - 2 sec^2(x) tan^2(x) + tan^4(x) 
1/cos4(x) - 2[1/cos2(x)][sin2(x)/cos2(x)] + [sin4(x)/cos4(x)]

[1-2sin2(x) + sin4(x)] / cos4(x)

[1 - sin2(x)]2 / cos4(x)

[cos2(x)]2 / cos4(x)

1

with trig identities, the trick is to write out all the terms using sin(x) and cos(x). Then start grouping like terms together (e.g., common denominator). Simplify compound terms whenever possible (e.g., sin²(x) + cos²(x) = 1), then repeat these steps as neccessary

[ResidentBiscuit]

with trig identities, the trick is to write out all the terms using sin(x) and cos(x)

Well hot damn, that makes some of these easier. I couldn't find this little trick anywhere. Got another one for you:

Perform given operations and simplify

[sin(x)/1+sin(x)] - [sin(x)/1-sin[x]

Ok so I thought this one was going to be easy. I got to common denominators, distributed each side, and I ended up with nothing left in the numerator. I got to

0/1-sin^2(x)

which is not an available answer.

[atropos]

[sin(x)/1+sin(x)] - [sin(x)/1-sin[x] 

{[sin(x) * (1 -sin(x)] / [(1+sin(x)) (1-sin(x))]}  - {[sin(x) * (1 +sin(x)] / [(1+sin(x)) (1-sin(x))]}

[sin(x)-sin2(x)-sin(x)-sin2(x)] / [1-sin2(x)]

-2sin2(x) / 1-sin2(x)

-2sin2(x) / cos2(x)

-2tan2(x)

[ResidentBiscuit]

Ok at line three, this is what I got:

[sin(x)-sin2(x)-sin(x)+sin2(x)] / [1-sin2(x)]

I guess this is where I got messed up. Didn't distribute the negative sign across. I distributed the sin(x) across, and then slapped a minus sign in there. Mind explaining why that's not going to yield the right answer?

[atropos]

Just focusing on the numerator of the second term here (my step 2):

-[sin(x) * (1+sin(x) )]
if you distribute the negative sign first, there are two options

1. -(sin(x) ) * (1+sin(x))  = -sin(x) - sin2(x)

2. sin(x) * -(1+sin(x) ) = sin(x) * (-1 -sin(x)) = -sin(x) - sin2(x)

if you distribute the sin first
-[sin(x) + sin2(x)] = -sin(x) - sin2(x)

So all roads lead to the same result. If I had to guess, I'd say that you forgot to completly distribute the negative sign on your step 3.

EDIT: or because of this

...and then slapped a minus sign in there

:P

[ResidentBiscuit]

or because of this
...and then slapped a minus sign in there
:P

Aah I figured that's probably what did it haha. It's been a good while since I've been in algebra class. I guess I've forgotten some random bits

[ResidentBiscuit]

Alright I have to say I'm understanding this much better now.

Thanks for your time, I appreciate it!

[atropos]

You're welcome.
It was actually fun for me.

yes, I'm a math geek