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?

