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Topic: Proving trigonometric identities
Replies: 32   Last Post: Jan 27, 2009 1:59 AM

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 Albert Posts: 282 Registered: 12/9/07
Re: Proving trigonometric identities
Posted: Jan 24, 2009 3:56 AM

On Jan 23, 9:21 pm, Angus Rodgers <twir...@bigfoot.com> wrote:
> On Fri, 23 Jan 2009 00:41:45 -0800 (PST), Albert
> >(tan x + sec x - 1) / (tan x - sec x + 1) = (1 + sin x) / cos x
> There may be a neater way to do it, but one fairly short way is
> to do the same first step as I suggested for the first problem,
> which is to rewrite A/B = C/D as AD = BC.  The left hand side
> of the latter identity simplifies to 1 + sin x - cos x.  Does
> the right hand side simplify to the same expression?

Yes - I did it just then as I got a bit bored of solving questions
like 4cos x - 3 = 0; for 0 degrees <= x <= 360 degrees.

The key is (like at the LHS) to make [the terms in the questions] in
terms of sin x and cos x. Once the RHS has a denominator of cos x,
then the last line of working is equal to what is on the LHS.

Back to solving equations:
Why does my booklet say 'none' to:
Problem 3. Solving the following equations for 0 degrees <= x <= 360
degrees.
(a) ...
(b) 3cosec x + 4 = 0;
(c)...
etc...

3cosec x = -4
cosec x = -4 / 3
cosec x is quotient of the hypotenuse divided by the opposite. Our
value here is negative and since the hypotenuse is always positive,
the opposite side must be negative and hence the answer will be in one
of the left quadrants (II or III). So then there are solutions to the
equation, aren't there? and why/why not?

Date Subject Author
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Angus Rodgers
1/20/09 Guest
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 matt271829-news@yahoo.co.uk
1/21/09 Albert
1/21/09 matt271829-news@yahoo.co.uk
1/23/09 Albert
1/23/09 Angus Rodgers
1/23/09 Angus Rodgers
1/23/09 Passerby
1/23/09 Dave Dodson
1/24/09 Albert
1/24/09 Angus Rodgers
1/24/09 Albert
1/26/09 Albert
1/26/09 Driveby
1/26/09 Albert
1/26/09 A N Niel
1/27/09 Albert