Verifying Trig Identities

[mynamesmurph]

Can you help me verify this identity?

ln|secx| = -ln|cosx|


Most of the problems I had were nothing like this, they were something like, (tanx+cotx)/cscx=1/cosx

These weren't too bad, but I'm stumped now. Any help would be tremendous!

 

[HallsofIvy]

If you know about "ln(x)" then you should know about its inverse function, \(\displaystyle e^x\). What do you get if you take the exponential to both sides?

 

[mynamesmurph]

I'm a little rusty on logarithms, but I think I get what you're saying... But the instructions are to algebraically manipulate the left side to turn it to the right side. Maybe that's what you're leading me to, but I still don't understand.

 

[Ishuda]

ln(xⁿ) = n ln(x)
What is the relationship between sec(x) and cos(x)?

 

[mynamesmurph]

ln(xⁿ) = n ln(x)
What is the relationship between sec(x) and cos(x)?

Reciprocal identities.

 

[mynamesmurph]

I understand the logarithm equality too, I must be missing something obvious.

 

[HallsofIvy]

Are you possibly missing the fact that -ln(x)= ln(x^-1)= ln(1/x)?
From ln(|sec(x)|)= -ln(|cos(x)|)= ln(|1/cos(x)|), taking the "exponential" or taking the inverse function, or just using the fact that ln is a "one-to-one" function, we have immediately that |sec(x)|= |1/cos(x)|. Do you see what "reciprocal identity" mynamesmurph meant?

 

[mynamesmurph]

OK, I think I got it.

ln|secx|
ln|1/cosx|
ln|(cosx)^-1|
-ln|cosx|

That should do it, right?

 

[Ishuda]

OK, I think I got it.

ln|secx|
ln|1/cosx|
ln|(cosx)^-1|
-ln|cosx|

That should do it, right?

Yep.

 

[mynamesmurph]

Much thanks, you guys are great.