Help with Secant Proof

[MGRGRM]

Hi, I've tried a few methods for this proof (such as writing the left side in terms of cosine and the right side with tangent), but I'm not getting anywhere, so some help to start me off would be greatly appreciated. Are there any identities I can use to help prove this? Thank you!

A picture of the proof written out is included

 

[skeeter]

Hi, I've tried a few methods for this proof (such as writing the left side in terms of cosine and the right side with tangent), but I'm not getting anywhere, so some help to start me off would be greatly appreciated. Are there any identities I can use to help prove this? Thank you!

A picture of the proof written out is included
View attachment 26370

this identity may help ...

[MATH]\cos^2{x} = \frac{1+\cos(2x)}{2}[/MATH]

 

[MGRGRM]

this identity may help ...

[MATH]\cos^2{x} = \frac{1+\cos(2x)}{2}[/MATH]

Thank you! I'm still a little confused about how to apply that identity, but I'll try some things out and hopefully, it'll work

 

[nasi112]

[MATH]2\sec 2x = \frac{2}{\cos 2x} = \frac{2}{2\cos^2 x - 1} = \frac{2}{\frac{2}{\sec^2 x} - \frac{\sec^2 x}{\sec^2 x}} = \frac{2\sec^2 x}{2 - \sec^2 x}[/MATH]
Can you continue?

 

[MGRGRM]

[MATH]2\sec 2x = \frac{2}{\cos 2x} = \frac{2}{2\cos^2 x - 1} = \frac{2}{\frac{2}{\sec^2 x} - \frac{\sec^2 x}{\sec^2 x}} = \frac{2\sec^2 x}{2 - \sec^2 x}[/MATH]
Can you continue?

Thank you so much for the help! I feel kind of dumb because you've given me a really good head start, but I really can't figure out what to do next. I've been trying some things, but I don't know how I can clear the denominator out in a way that's legal for proofs that'll then allow me to work on what's in the numerator. I'll spend some more time on it and hopefully, I'll get somewhere.

 

[JeffM]

The secant does not exist if the cosine is zero so the proof must involve, explicitly or implicitly, some restriction on x. Given such a restriction, what skeeter was hoping you would find is

[MATH]cos^2(x) = \dfrac{1 + cos(2x)}{2} \implies \dfrac{1}{cos^2(x)} = \dfrac{2}{1 + cos(2x)} \implies sec^2(x) = \dfrac{2}{1 + cos(2x)} \implies WHAT?[/MATH]
Now if you do not happen to have skeeter’s identity at your finger tips, you will need the double angle formula.

[MATH]sec^2(x) = \dfrac{1}{cos^2(x)} = \dfrac{2}{2 cos^2} = \dfrac{1}{2cos^2(x) - 1 + 1} = \dfrac{2}{1 + cos(2x)}.[/MATH]

 

[JeffM]

Thank you so much for the help! I feel kind of dumb because you've given me a really good head start, but I really can't figure out what to do next. I've been trying some things, but I don't know how I can clear the denominator out in a way that's legal for proofs that'll then allow me to work on what's in the numerator. I'll spend some more time on it and hopefully, I'll get somewhere.

If you followed my post above then

[MATH]sec^2(x) = \dfrac{2}{1 + \dfrac{1}{sec(2x)}} = WHAT?[/MATH]

 

[MGRGRM]

The secant does not exist if the cosine is zero so the proof must involve, explicitly or implicitly, some restriction on x. Given such a restriction, what skeeter was hoping you would find is

[MATH]cos^2(x) = \dfrac{1 + cos(2x)}{2} \implies \dfrac{1}{cos^2(x)} = \dfrac{2}{1 + cos(2x)} \implies sec^2(x) = \dfrac{2}{1 + cos(2x)} \implies WHAT?[/MATH]
Now if you do not happen to have skeeter’s identity at your finger tips, you will need the double angle formula.

[MATH]sec^2(x) = \dfrac{1}{cos^2(x)} = \dfrac{2}{2 cos^2} = \dfrac{1}{2cos^2(x) - 1 + 1} = \dfrac{2}{1 + cos(2x)}.[/MATH]

Thank you! I understand this identity now (I hope), but I'm still unsure about how all of this fits into my proof. As in, how should I get from 2sec(2x) to where I can apply that identity? Right now, the step I'm on is where nasi112 left off, which is 2sec²x/2-sec²x, so can I use that identity going forward?

 

[lex]

@MGRGRM

skeeter's hint was:
[MATH]\cos^2{x} = \frac{1+\cos2x}{2}[/MATH]so [MATH]\frac{1}{\cos^2 x} = \frac{2}{(1+\cos2x)}\quad (1)[/MATH]
You want to prove that: [MATH]\quad 2\sec2x=\sec^2{x}(1+\sec2x) [/MATH]
[MATH]RHS=\frac{1}{\cos^2{x}}\left(1+\frac{1}{\cos2x}\right)[/MATH]
[MATH]=\frac{1}{\cos^2{x}}\left(\frac{1+\cos2x}{\cos2x}\right)[/MATH]
Now use (1) above ...

 

[MGRGRM]

@MGRGRM

skeeter's hint was:
[MATH]\cos^2{x} = \frac{1+\cos2x}{2}[/MATH]so [MATH]\frac{1}{\cos^2 x} = \frac{2}{(1+\cos2x)}\quad (1)[/MATH]
You want to prove that: [MATH]\quad 2\sec2x=\sec^2{x}(1+\sec2x) [/MATH]
[MATH]RHS=\frac{1}{\cos^2{x}}\left(1+\frac{1}{\cos2x}\right)[/MATH]
[MATH]=\frac{1}{\cos^2{x}}\left(\frac{1+\cos2x}{\cos2x}\right)[/MATH]
Now use (1) above ...

Thank you so much! I think I got it now!

 

[lex]

Great. Do you want to post your solution? It's up to you.

 

[lex]

For completeness:

[MATH]\cos^2{x} = \frac{1+\cos2x}{2}[/MATH]so [MATH]\frac{1}{\cos^2 x} = \frac{2}{(1+\cos2x)}\quad (1)[/MATH](when [MATH]\text{cos}x ≠ 0[/MATH])

You want to prove that: [MATH]\quad \boxed{ \hspace{0.5cm} 2\sec2x=\sec^2{x}(1+\sec2x) \hspace{0.5cm}} [/MATH]
[MATH]RHS=\frac{1}{\cos^2{x}}\left(1+\frac{1}{\cos2x}\right)[/MATH]
[MATH]=\frac{1}{\cos^2{x}}\left(\frac{1+\cos2x}{\cos2x}\right)[/MATH]
[MATH]=\frac{2}{1+\cos2x}\left(\frac{1+\cos2x}{\cos2x}\right)[/MATH] by (1) above

[MATH]=\frac{2}{\cos2x}[/MATH]
[MATH]=2\sec2x[/MATH]
[MATH]=LHS[/MATH]
[MATH]\therefore \quad 2\sec2x=\sec^2{x}(1+\sec2x) \quad \text{Q.E.D.}[/MATH]