Question

Trig Identities

Answer 1

\[(\sin^4x-\cos^4x)/ (\sin^2x -\cos^2x) \]

Answer 2

i have to make it equal to 1

Answer 3

Oh this is a fun one :3

Answer 4

I'm not good at proving Identities ;/

Answer 5

Ya this one is a bit tricky D:
Let's work on the numerator a sec.

Answer 6

Think about it in this form for a moment,
\[\Large a^4-b^4\]We can rewrite the powers like this,\[\Large (a^2)^2-(b^2)^2\]

Do you remember how to deal with the `difference of squares`?
They can be written as the product of conjugates like this,\[\Large x^2-y^2=(x-y)(x+y)\]

Answer 7

yeah

Answer 8

Applying this to our a and b equation gives us,\[\Large (a^2)^2-(b^2)^2 \qquad=\qquad (a^2-b^2)(a^2+b^2)\]

Answer 9

There are a ton of powers floating around :D make sure you understand that part a sec.

Answer 10

so it becomes :\[(\sin^2x-\cos^2x)(\sin^2x-\cos^2x)\]

Answer 11

Woops, should be `addition` in one of those brackets.

Answer 12

oh oops

Answer 13

then i cross it out to get sin^2x+cos^2x

Answer 14

and thats 1

Answer 15

Yay good job \c:/

Answer 16

ahh i see

Answer 17

I'm still not to good on proving trig functions do you have any tips

Answer 18

Manipulating Trig Functions takes soooo much practice :(
I'm not sure really want to say besides `do tons of problems`.

There are so many identities. At times, it can be tough to know what tool to grab from your math goodie bag to solve a problem.

Answer 19

I mean there are a few general tips I could give...
Usually when you see squares, using your `square identities` will help.

I dunno, every problem is different :3 it's hard to say what path to take.

Answer 20

I can second that. I'm sure there are some proof problems that we might have to look at and manipulate for a second before we see it. Identities are really just practice and getting better and better at recognition. You also start to know what to look for. If you can't see anything obvious you learn to maybe multiply by conjugates, change things to sines and cosines, recognizing if there might be a quadratic or some sort of factoring hidden (see your above problem). It's just stuff you get better at seeing. It won't be easy for a while, but you'll getit ^_^

Answer 21

ah man i dont really have much time on my hands. I'll try but thank you

Answer 22

And to be honest, I think I sucked at identities when i was done with trig. It was doing more and more math after that where I learned more tricks and started recognizing things. The idea is don't give up and don't feel bad if you don't get it, these are hard. Just do what you can and we're here to help :P

Answer 23

Yah I feel bad for students that take trig, it really demands a lot of time from you. Which can be tough if you're a busy gal :c

Answer 24

alright thanks a bunch! im just flippin out for my final tomorrow

Answer 25

Yeah, and my professor went out of his way to make the class difficult -_- He explained very well then gave demonic exams :/

Answer 26

when proving a trig function would you try to change things to sin and cos first?

Answer 27

I think sine and cosine look easier on the eyes for people at first. Sometimes it makes things a little more obvious. Anyway, the identities we start to memorize have to do with sines and cosines, right? I remember sin^2(x) + cos^2(x) = 1. But I still have to mentally divide everything by sin^2(x) and think for a second before I pull out 1+ cot^2(x) = csc^2(x). Until you get better at it, a lot of the time sines and cosines just let you see things better.

Answer 28

your final is tomorrow? just keep posting questions then!! :O
and keep studying! :D

Answer 29

yeah ):

Answer 30

ya when its all in sin and cos its easier to think for some reason

Answer 31

\[(1-sinx)/(cosx) = (cosx)/(1+sinx)\]

Answer 32

Looks like a conjugate one.

Answer 33

corss multiply u will see magic

Answer 34

i have to prove it

Answer 35

Well, they teach you to only deal with one side xDD They don't like you going back and forth or mixing sides because the you're not technically "proving" it.