Proof

[Joanna]

Would anyone have a hint for me or help me start this proof:
Sec^2(x/2)= (2secx+2)/(secx+2+ cosx)

 

[mmm4444bot]

I would start by rewriting the secants in terms of cosine, and simplifying.

 

[Joanna]

So if I start with the right side right? Then I get: [2(1/cosx) +2 / (1/cosx) + 2 + cosx ]
So then does the (1/cosx) and the +2 cancel out leaving 2/cosx ???

 

[mmm4444bot]

Before continuing, I would like to know whether this is one of those proofs where your instructor restricts you to working on one side only because that's not what I had in mind. There are different proofs.

 

[mmm4444bot]

So if I start with the right side right?

I'm not sure what you're asking.

Then I get: [2(1/cosx) +2 ]/[ (1/cosx) + 2 + cosx ]

So then does the (1/cosx) and the +2 cancel out leaving 2/cosx ?

oh, not at all!

It's a compound algebraic ratio that needs to be simplified to a single ratio (i.e., no fractions in either the numerator or denominator) by combining terms and using the famous relationship: A/B divided by C/D = A/B times D/C.

In other words, dividing by a fraction is the same as multiplying by its reciprocal.

PS: I inserted the missing brackets (in red). :cool:

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Do you know how to combine the terms 1/cos(x) + cos(x) + 2 into a single ratio?

 

[mmm4444bot]

Okay, my proof is ready.

I replaced the secants on each side with the respective reciprocal of cosines, and then I simplified the result on each side to arrive at the same simple ratio:

2/[cos(x) + 1]

Oh, yeah, before simplifying the lefthand side, I also used the power-reducing identity for cosine squared:

cos(u)*cos(u) = [cos(2*u) + 1]/2

 

[Joanna]

Ok so I combined everything to a single fraction and multiplied it by the reciprocal to get 2/(1+ cos^2x) but how do I get from there to sec^2(x/2)
My teacher told us we need to show that one side is equal to the other, we can't just work on them together and show that they meet somewhere in the middle- he would not accept that :/

 

[mmm4444bot]

I combined everything to a single fraction and multiplied it by the reciprocal to get 2/(1+ cos^2x)

How did you get cosine squared?

but how do I get from there to sec^2(x/2)

Simplify from sec(x/2)^2 to 2/[cos(x) + 1] first (using the identity that I gave you); then, you will see the steps going backwards from 2/[cos(x) + 1] to sec(x/2)^2.

My teacher told us we need to show that one side is equal to the other

These proofs can be easily done, working on both sides simultaneously.

See "Transitive Property of Equality".

we can't just work on them together and show that they meet somewhere in the middle- he would not accept that

Seems like a pointless objection, as we can always read down one side and up the other.

So, to comply, write the steps for the lefthand side (in reverse, as I described above) directly below your corrected simplification of the righthand side.

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Are you able to use the power-reducing identity for cos(u)^2 and simplifications to go from 1/cos(x/2)^2 to 2/[cos(x) + 1] ?

 

[Joanna]

this is frustrating because I cannot see how you got from Sec^2(x/2) to cos(u)*cos(u) = [cos(2*u) + 1]/2

and i got 2/ 1 + cos^2x because:

2secx+ 2/ secx + 2 +cosx = (2/cosx) + 2 / (1/cosx) + 2 + cosx (this is where i multiply to get the same denominator ) = [ (2 + 2cosx/ cosx) / (1 + 2cosx + cos^2/ cosx) = (2 + 2cosx/cox) * (cosx/1 + 2cosx+ cos^2x) things cross out and you get 2/ 1+ cos^2x

 

[mmm4444bot]

I cannot see how you got from Sec^2(x/2) to cos(u)*cos(u) = [cos(2*u) + 1]/2

That's because I didn't.



and i got 2/[1 + cos^2x] because:

2secx+ 2/ secx + 2 +cosx =

(2/cosx) + 2 / (1/cosx) + 2 + cosx =

[ (2 + 2cosx/ cosx) / (1 + 2cosx + cos^2/ cosx) =

(2 + 2cosx/cox) * (cosx/1 + 2cosx+ cos^2x)

things cross out and you get 2/ 1+ cos^2x

"Things" do not cross out like that!

You must learn the basic algebra of fractions, before you try to learn trigonometry.

(a + b)/a does not allow us to cross out the a's simply because we see one on top and bottom

We can only do that when we have a*b on top, instead of a + b

In other words, a*b/a allows us to simply cross out the a's

(This is the third time that you've confused factors for addends)

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It's hard for me to decipher a lot of your typing above, but I think that you're okay up to the result:

[2 + 2*cos(x)]/[1 + 2*cos(x) + cos(x)^2]

Factor both the numerator and the denominator; only then will you have common factors to cancel.

The denominator has this form:

u^2 + 2*u + 1

where symbol u represents cos(x).

Can you factor u^2 + 2*u + 1 ?

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As for the left-hand side, please show me what you did when you rewrote sec^2(x/2) in terms of cosine.