Trig Identities: Proof

M

Monkeyseat

Full Member
Joined
Jul 3, 2005
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Sorry for the scan but I don't know how to write this out.

I am stuck on question (j), I don't know where to start...

trigeq6.png


I think I have to use:

tanon6.png


But still don't know how to approach this. Sorry for the lack of working.

Any clues?

Thanks.
 
\(\displaystyle \tan (x)\sin (x) = \frac{{\sin ^2 (x)}}{{\cos (x)}} = \frac{{1 - \cos ^2 }}{{\cos (x)}} = \frac{{\left( {1 - \cos } \right)\left( {1 + \cos } \right)}}{{\cos (x)}}\)
 
pka said:
\(\displaystyle \tan (x)\sin (x) = \frac{{\sin ^2 (x)}}{{\cos (x)}} = \frac{{1 - \cos ^2 }}{{\cos (x)}} = \frac{{\left( {1 - \cos } \right)\left( {1 + \cos } \right)}}{{\cos (x)}}\)
Click to expand...

Should "1 - cos^2" not be "1 - cos^2 x"? I don't know what happened to the "1 - cosx" that was on the denominator in the question...

Ok, so I've got:

(1-cosx)/(cosx) * (1+cosx)/(cosx) = 1 + (1/cosx)

Would the left side all be over "1 - cosx"?

I really don't understand these. Sorry, I don't want it to seem like I'm just coming here to get the answers, but I only started doing these today and quite get my head around them...
 
Sorry, I've been looking at this for nearly 2 hours and I am still stumped on where to go.

If you wouldn't mind, please could you give me a bit more help?

Thanks.

I got:

((1-cos^2 x)(1+cos^2 x)/cosx)/1-cosx

No idea how to sort that out.
 
\(\displaystyle \frac{{\overbrace {1 - \cos ^2 (x)}^{\mbox{difference of squares}}}}{{\cos (x)\left[ {1 - \cos (x)} \right]}} =\frac{{\left[ {1 - \cos (x)} \right]\left[ {1 + \cos (x)} \right]}}{{\cos (x)\left[ {1 - \cos (x)}\right]}} = \frac{{\left[ {1 + \cos (x)} \right]}}{{\cos (x)}}\)
 
Ok I think I might have it now:

cosmw2.png


Is that correct?

The thing that threw me was when (x) and the original denominator disappeared in your first post. I didn't know the original "1 - cosx" denominator and "cosx" denominator from sinx/cosx went on the same line/demoninator (is that right?). If that makes sense... I was originally doing it in 2 divisions like: (1-cosx)/(cosx) * (1+cosx)/(cosx) = 1 + (1/cosx).

Thanks for helping. I meant (x) and theta as the same thing, (x) is just easier to type as text for me on the computer.
 
Yes, by George I think you have.

You know that you should learn to use LaTeX.
 
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