Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

#1 2006-01-07 10:46:35

number0001
Member
Registered: 2006-01-06
Posts: 3

Help--Another Trig Problem

I got another trig problem, which is to prove (make the left side of the "=" equal to the right side).

(cos3x-cosx)/(sin3x-sinx)=-tan2x

I tried it but got stuck. Thanks very much for any help!

Offline

#2 2006-01-07 11:00:11

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Help--Another Trig Problem

I could work it out myself but I think you should.

First I'd suggest replacing cos(3x) and sin(3x) with cos(2x + x) and sin(2x + x) then use the sum identities. We did this in the previous example. And note the identity tan (2x) = (2 tan x)/(1 - tan^2 (x) ). Also note that tan x = (sin x)/(cos x)

You can do it!

A logarithm is just a misspelled algorithm.

Offline

#3 2006-01-07 11:03:08

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Help--Another Trig Problem

Also note the identiy for sin(2x) and the three identities for cos(2x).

I'd really really really reccomend printing out a reference sheet of all the indentities. I have a binder of lots of difference indentities and math notes. This is math, not spelling. Understanding why a formula works is more important then remembering it. Just print out a reference sheet.

A logarithm is just a misspelled algorithm.

Offline

#4 2006-01-07 18:31:07

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Help--Another Trig Problem

If you really can't get it I'll explain how to do it, but you really ought to learn to do this for yourself.  I wouldn't really be helping you if I just did the problem for you.

A logarithm is just a misspelled algorithm.

Offline