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LynnM · Aug 20, 2010, 12:20 PM

I need to manipulate one side so that it becomes the same as the other side. I've tried to manipulate either side but I can't seem to figure it out. Step by step PLEASE!

Unknown008 · Aug 20, 2010, 12:57 PM

I think I'll use the right side to prove the left side. I'll be using 'A' instead of 'x' so that it's easier to locate the variable, instead of putting (x) each time.

$\frac{2cosA-2cos^2A}{2sinAcosA} = \frac{2cosA(1-cosA)}{2sinAcosA}$

$= \frac{\cancel{2cosA}(1-cosA)}{\cancel{2}sinA\cancel{cosA}}$

$= \frac{1-cosA}{sinA}$

$= \frac{1-cosA}{sinA} \times \frac{\frac{1}{cosA}}{\frac{1}{cosA}}$

$= \frac{secA - 1}{tanA} \times \frac{secA+1}{secA+1}$

$= \frac{sec^2A - 1}{tanA(secA+1)}$

$= \frac{tan^2A}{tanA(secA+1)}$

$= \frac{tanA}{secA+1}$

:)

LynnM · Aug 20, 2010, 01:30 PM

Thank you SOOO much for your help! I am doing grade 12 math online so it's kind of difficult to understand everything without the instructor there to explain it. I was wondering why you multiply all by (1/cosA)/(1/cosA). I've read about multiplying by the conjugate so is this something similar? I ask because I would have multiplied by the conjugate which is apparently wrong... right?

galactus · Aug 20, 2010, 07:05 PM

Yes, you can do what the U Man done, but be sure to do it to the top and bottom. You can do all sorts of things in math, just make sure it's balanced.

Here is another approach starting with the left side and working toward the right.

$\frac{tan(x)}{sec(x)+1}$

$\frac{\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}+1}$

$\frac{\frac{sin(x)}{cos(x)}}{\frac{1+cos(x)}{cos(x )}}$

$\frac{sin(x)}{cos(x)}\cdot\frac{cos(x)}{1+cos(x)}$

$\frac{sin(x)}{(1+cos(x))}\cdot\frac{(1-cos(x))}{(1-cos(x))}$

$\frac{sin(x)(1-cos(x))}{sin^{2}(x)}$

$\frac{1-cos(x)}{sin(x)}$

Multiply top and bottom by
2cos(x):

$\frac{2cos(x)(1-cos(x))}{2cos(x)sin(x)}$

$\frac{2cos(x)-2cos^{2}(x)}{2sin(x)cos(x)}$

See? There is more than one way to skin a trigonometric identity.:):)

LynnM · Aug 20, 2010, 07:35 PM

Oh, wow! For the million and one times I've tried to prove this, I have never come up with anything even close to what you two have shown me. So, the "Multiply top and bottom by 2cos(x)", how did you know to do this? Because the RHS has this? I can't figure out how you know to do steps like these.

Thanks for all of your help. I might still be a little lost but I'm definitely getting there... I think! Lol!

Unknown008 · Aug 20, 2010, 10:46 PM

What you need is to reach your objective, and for that, you can multiply by 1, right? You can see that anything that we used to multiply with the numerator and the denominator would otherwise cancel out as being 1.

We do this so that some functions cancel out, or new functions that is closer to the 'objective' emerge.

For example, the first thing that I thought of is removing the '2' so that I get the '1' in the answer. From there, I needed to get sec and tan, and remove this cos. So, I divided everything by cos. Then, I needed to have sec in the denominator and remove the sec in the numerator. 'Rationalising' using secA + 1 will remove the sec in the numerator and introduce it in the denominator. Then, things were easy :)

As for galactus, he needed to introduce the '2', once things were much simpler. (And by the way, it's a good idea to have everything in cos and sin in proofs to simplify things). Then introduce what was missing.

I hope it helped! :)