But Sec(T) is not equal to Cosh(T), for instance:
Sec(Pi) = -1,
Cosh(Pi) = 11.59
Are you sure you have written down the problem correctly?
Hey all,
I'm supposed to show that Sec (Theta) = Cosh (Theta)
Given:
x = ln (Sec (theta) + tan (theta))
I have NO idea where to start, any pointers?
-B
P.S.
I'm supposed to compute the integral of x^5(Sin(x^3)). I should use parts to integrate this, correct?
I thought it had something to do with the ln function, possibly dealing with the derivatives of the trigonometric functions as they related to a fraction, which would eliminate the trig function in and of itself and leave the fraction with only constants and x terms...(Hope that made sense)
At any rate, I'll ask the prof tomorrow to clarify, maybe he miswrote the problem...
Thanks,
-B
Did you simply want a value for which Sec(T) = Cosh(t)? There are an infinite number of them, you can show that
but remember that the complex natural log is a multivalued function. In putting it in this form I may have even gotten rid of some solutions... anyone have some pointers on finding all solutions?
I was able to clarify what my prof wanted for the sec = cosh problem:
He wanted us to take the value for cosh (having to do with the e^x terms)
and relate it to the ln function (as it pertains to e^x terms)
Hope that makes sense...
At any rate, I tried plugging the values into the cosh function, but didn't get an answer or I didn't get an answer that made the problem look done...
Any thoughts?
-B
Ok i figured it out. You did write it down slightly wrong. What he wants is for you to show Cosh(x) = Sec(T).
You can raise e by both sides of your equation. Then you know e^x. You also know e^-x. Then add them. You can solve it down to Cosh(x) = Sec(T).
For example if we had:
x = ln (y)
Then raise e to the power of both sides:
e^x = e ^ [ln (y)] = y
This is the same as saying:
If 2 = 2 then e^2 = e^2. Right?
Doing this is often used if you have a natural log since it is the inverse natural log:
e^[ln (x)] = x