<p>Yeah i know the SAT doesn't test this but I have to do this problem and I know a bunch of braniacs are on this site who can probably solve this:</p>
<p>Prove: </p>
<p>(1 + cosx + sinx)/(1 + cosx - sinx) = tanx + secx</p>
<p>It might be a lot easier if you write it out. </p>
<p>thanks in advance</p>
<p>Convert tanx into (sinx / cosx) and secx into (1 / cosx). </p>
<p>Now you will have two fractions set equal to each other...</p>
<p>(1 + cosx + sinx)/(1 + cosx - sinx) = (sinx + 1) / (cosx)</p>
<p>so cross multiply. You should get:</p>
<p>cosx + (cosx)^2 + sinxcosx = 1 + sinxcosx + cosx - (sinx)^2 - sinx + sinx</p>
<p>The sinx on the right side cancels, the sinxcosx cancels on each side, the cosx cancels on each side. Add the (sinx)^2 to the left side and you have:</p>
<p>(cosx)^2 + (sinx)^2 = 1</p>
<p>Since the left side indeed equals 1, you have an identity and have proven equality.</p>
<p>(1 + cosx + sinx)/(1 + cosx - sinx) = (sinx + 1) / cosx</p>
<p>(1 + cosx + sinx) * cosx = (1 + cosx - sinx) * (sinx + 1)</p>
<p>cosx + cos^2(x) + sinx<em>cosx = sinx + sinx</em>cosx - sin^2(x)+1+cosx-sinx</p>
<p>(eliminate the same trig from the both sides)</p>
<p>cos^2(x) + sin^2(x) = 1 (idk what you call it, but we say a basic equality).</p>
<p>so it means every x can satisfy the equation except (1 + cosx - sinx) = 0 and cosx = 0. Because they're below (still dunno proper term).</p>
<p>it could be either wrong or circular approach.</p>
<p>Oh actually for this problem it requires that you work it out from the left side not the right, it's only acceptable if you work from the left side of the equation...can you do it from the left please? i could do it from the right but don't get how to do it from the left.</p>
<p>Well you're not really doing it from either side as you end up doing cross multiplication. That's the easiest way. You could try splitting up the fraction on the left but I don't think it's possible to complete the problem solely by manipulating the left side of the equation.</p>
<p>Well the thing is, the original problem was (1+tan1/2x)/(1-tan1/2x) = tanx + secx</p>
<p>I simplified the left all the way to (1 + cosx + sinx)/(1 + cosx - sinx) and couldn't get any further. I don't see how you can work it from the left side in the original problem (on the first line of this post).</p>
<p>The way people have been doing it isn't exactly rigorous. You can't start with something and show that it leads to something true and expect that to prove your statement.</p>
<p>For example:</p>
<p>1 = 2
Multiply by 0.
0 = 0
Therefore, 1=2</p>
<p>You have to start with something true and then get to what you're trying to prove.</p>
<p>Let u = (1+cosx), u^2 = 1 + cos^2x+2cosx</p>
<p>[ (1+cosx) + sinx ] / [ (1+cosx) - sinx ]
=[ u + sinx ] / [ u - sinx ]
= [ u + sinx ] / [ u - sinx ] * [u + sinx]/[u + sinx]
= [u^2 + sin^2x + 2usinx ] / [u^2 - sin^2x]
= [1 + cos^2x + 2cosx + sin^2x + 2(1+cosx)(sinx)] / [1 + cos^2x + 2cosx - sin^2x]
= [2+2cosx + 2 sinx + 2sinxcosx] / [2-sin^2x + 2cosx - sin^2x]
= [1+cosx + sinx + sinxcosx] / [1-sin^2x + cosx]
= [1+cosx + sinx + sinxcosx] / [cos^2x + cosx]
= [(cosx+1)(1+sinx)] / [cosx(cosx+1)]
= [1+sinx] / [cosx]
= tanx + secx</p>
<p>There's probably a better way to do it, but I'm lazy.</p>
<p>I'd imagine that the high school course isn't looking for a proof in the sense that you're implying. ;) Most kids would flunk.</p>
<p>wow i give up...</p>
<p>This is part of the required math syllabus for 10th graders here in India.</p>
<p>hey dude, you realize you're hijacking my thread?</p>
<p>Yeah...sorry, want me to delete that post?</p>
<p>Not that I meant to hijack though, I was just posting some food for thought (or so I felt)</p>
<p>
[Quote]
I'd imagine that the high school course isn't looking for a proof in the sense that you're implying.
[/Quote]
</p>
<p>Ummm... as opposed to what? Sorry.....</p>
<p>LAst year i had two tests on these friggin trig identities and from what i remember(100 both tests ) your not allowed to cross multiply these problems.</p>
<p>There are three very useful identities, which for some reason are not commonly presented in a high school trig course.
They are based on the universal substitution u = tan(x/2):</p>
<p>sin(x) = 2u / (1 + u^2)</p>
<p>cos(x) = (1 - u^2) / (1 + u^2)</p>
<p>tan(x) = 2u / (1 – u^2).</p>
<p>Going to the original problem:
(1 + tan(x/2)) / (1 – tan(x/2)) = </p>
<p>(1 + u) / (1 – u) = </p>
<p>[(1 + u) * (1 + u)] / [(1 - u) *(1 + u)] = </p>
<p>( 1 + 2u + u^2) / (1 – u^2) = </p>
<p>2u / (1 – u^2) + (1 + u^2) / (1 – u^2) =</p>
<p>tan(x) + 1/cos(x) = </p>
<p>tan(x) + sec(x).</p>
<p>Are you sure universal substitution must be used here? And it's a precalculus course actually at a community college, not in high school.</p>
<p>
[quote]
Ummm... as opposed to what? Sorry.....
[/quote]
</p>
<p>You're making it sound like it needs to be proven. As in not using the technique given, but proving it beyond all doubt. That's typically reserved for college level courses (Real Analysis, Introduction to Higher Mathematics, whatever your school may call it). For a high school class, proving that one side equals the other via simple trigonometric identities (as I did in my first post) will typically suffice. Then again, there are of course some high schools that may require more rigorous proofs, though for a precal coures at a CC, I can guarantee that my method would suffice.</p>
<p>
There us always more than one way of solving/proving something. brand_182 and Tsenguun gave one proof, mathwiz another, and I yet another.</p>
<p>My proof, IMMO, is the easiest one, but I cant say that it is the only right one.</p>
<p> [quote] Well the thing is, the original problem was (1+tan1/2x)/(1-tan1/2x) = tanx + secx
tan(x/2) is already present in this problem, which leads me to believe that it was designed to have the universal substitution implemented in the proof.
You probably can find the universal substitution formulas in your precalculus course.</p>
<p>Two more cents.
Actually, you can start with something and show that it leads to something true, and that will prove the original statement. There is fine print though: all the transformations must be identical.</p>
<p>You can skip the explanations and examples and go straight to the end of this post.</p>
<p>==================== If x = y, then x<em>c = y</em>c, and another way around, x<em>c = y</em>c -> x = y, if c /= 0 (/ stands for “not”). Multiplying by non-zero is reversible.</p>
<p>Multiplying both x and y by 0 is not reversible, because you can’t divide by 0, therefore x<em>0 = y</em>0 /-> x = y.</p>
<p> [quote] 1 = 2 Multiply by 0. 0 = 0 Therefore, 1=2
</p>
<p>A common name for reversible transformations is identical transformations.
You can use non-identical transformations for solving equations, but you always have to check your answers.</p>
<p>One typical non-identical transformation is squaring both sides of an equation.
This operation is not reversible, because
a = b -> a^2 = b^2, but
a^2 = b^2 /-> a = b.
For example, 1^2 = (-1)^2…</p>
<p>You can’t prove 1 = -1 by squaring both sides, just as you can’t prove this “identity by multiplying both sides by 0. :(</p>
<p>Back to squaring:
Solving sqrt(x + 1) = x - 1:
x + 1 = x^2 – 2x + 1
x^2 – 3x = 0
x(x – 3) = 0
x=0 or x=3.
x=0 is the extraneous root, which crept in because of non-identical transformation.</p>
<h1>Let’s get back to the subject.</h1>
<p>brand_182 and Tsenguun applied identical transformations without explicitly stating that
(1 + cosx - sinx) /= 0
and
cosx /= 0.</p>
<p>Unfortunately, TruthfuLie (or his teacher?) set a limitation:
[quote]
this problem … requires that you work it out from the left side.
[/quote]
That makes mathwith and me the good guys. :D</p>