Trig. expressions = help.

[Wholewheat458]

I do not understand these problems;
I understand the trig basics (identities, inverses, and a bit more), but
I have'nt got a clue of where to start with these:
(sec 2 u - tan 2 u) / (cos 2 v + sin 2 v)

[ (sec y - tan y)(sec y + tan y)] / [sec y]

and this one: cos x - cos 3 x

Honestly, any help would be appreciated!! :shy:

[BoundByAxioms]

I do not understand these problems;
I understand the trig basics (identities, inverses, and a bit more), but
I have'nt got a clue of where to start with these:
(sec 2 u - tan 2 u) / (cos 2 v + sin 2 v)

[ (sec y - tan y)(sec y + tan y)] / [sec y]

and this one: cos x - cos 3 x

Honestly, any help would be appreciated!! :shy:

I'm assuming that you want to simplify these expressions?

1) Change everything into terms of sines and cosines.
2) Look over your identities and see if anything cancels out or if you can combine terms.

Can you make an attempt at that, and then see what happens from there?

Here's some help for the second one:
[sec(y)-tan(y)][sec(y)+tan(y)]=sec2y-tan2y. Recall that tan2y=sec2y-1. Where does that lead you now?

[Wholewheat458]

your clue for the second one helped tons!! ^^
I attempted the first one again, but i still think it's wrong
heres what i did:
Numerator turns out to be -sin x/cos2 x
and the denominator is = to 1 (identity)
so.. then i put the fractions in one layer and got: cos 2 x/ 1-sin x
i replaced cos 2 x with 1... oh, :blushing: it =s 1.. ^^ Hehehe!!
thank you!!
but i still have a question with: cos x - cos 3 x
how do you begin? do you split the cos 3 x??
or is there some identity that i am missing.. confused

[snipez90]

It's not hard if you write cos(3x) as cos(2x+x) and use the double angle formula.

[BoundByAxioms]

your clue for the second one helped tons!! ^^
I attempted the first one again, but i still think it's wrong
heres what i did:
Numerator turns out to be -sin x/cos2 x
and the denominator is = to 1 (identity)
so.. then i put the fractions in one layer and got: cos 2 x/ 1-sin x
i replaced cos 2 x with 1... oh, :blushing: it =s 1.. ^^ Hehehe!!
thank you!!
but i still have a question with: cos x - cos 3 x
how do you begin? do you split the cos 3 x??
or is there some identity that i am missing.. confused

No problem! I'm happy to be able to contribute meaningfully to physicsforums finally. I've still got a lot of math to learn myself! As for the third one, do exactly as snipez has suggested.