1. ## Evaluating an expression.

i have;

$\frac{(sec^2x - tan x)}{e^x}$, how do i calulate this to get 0.46??? if $x = \pi/4$

using; $\displaystyle \sec{x} = \frac{1}{\cos{x}}$;

$\frac{( \frac{(2)}{cos 2x +1} - tan x)}{e^x}$ whats next?

on a similar topic; i have;

$\frac{(sec^2x - tan x)}{e^x}$, how do i calulate this to get 0.46??? if $x = \pi/4$

using; $\displaystyle \sec{x} = \frac{1}{\cos{x}}$;

$\frac{( \frac{(2)}{cos 2x +1} - tan x)}{e^x}$ whats next?
Does the question ask you to find an answer correct to two decimal places? Then you substitute x = pi/4 into the expression and evaluate it using a calculator.

(Note: The exact value of each trig function should be known to you but are not required if an approximate answer is asked for).

I have no idea why you would be trying to re-write the expression.

3. I can seem to get the answer the book requires, when i put the numbers in to the formula?

I know that;
$\displaystyle \sec{x} = \frac{1}{\cos{x}}$

but what is $sec^2(x)$

I can seem to get the answer the book requires, when i put the numbers in to the formula?

I know that;
$\displaystyle \sec{x} = \frac{1}{\cos{x}}$

but what is $sec^2(x)$
ITs ok, i think ive got it now!

$sec^2(x)= \displaystyle \frac{1}{\cos{x}^2}$

therefore does;

$sec^2(x^2)= \displaystyle \frac{2}{\cos{2(x^2)}+1}$???

5. No, it's nothing like that. since $sec(x)= \frac{1}{cos(x)}$, $sec^2(x)= \left(\frac{1}{cos(x)}\right)^2= \frac{1}{cos^2(x)}$. You square the entire function, not just "x".

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

mr fantastic suggested that you put the numbers into a calculator. Is the problem that your calculator does not have a "sec" key? If so then $sec(x)= \frac{1}{cos(x)}$ means that you can use the "cos" key followed by the "1/x" key:
If $x= \pi/4$ then cos(x)= 0.70710678118654752440084436210485 and the reciprocal (1/x) of that is 1.4142135623730950488016887242097. Pressing the " $x^2$" key now gives 2. That should be no surprise. One of the "standard" values of cosine you should memorize is $cos(\pi/4)= \frac{\sqrt{2}}{2}$. Inverting that gives $sec(\pi/4)= \frac{2}{\sqrt{2}}= \sqrt{2}$ and, squaring, 2.

Of course, since $\pi/4= \pi/2- \pi/4$, $sin(\pi/4)= cos(\pi/4)= \frac{\sqrt{2}}{2}$ also so that $tan(\pi/4)= sin(\pi/4)/cos(\pi/4)= \left(\sqrt{2}/2}\right)/\left(\sqrt{2}/2\right)= 1$. $sec^2(\pi/4)- tan(\pi/4)= 2- 1= 1$.

Now, what is $e^{\pi/4}$? What is $\frac{1}{e^{\pi/4}}$?

6. so from all of this we gather;

1. $sec(x)= \frac{1}{cos(x)}$

2. $sec(x^2)= \frac{1}{cos(x^2)}$

3. $sec^2(x)= \left(\frac{1}{cos(x)}\right)^2= \frac{1}{cos^2(x)}$

4. $sec^2(x^2)= \left(\frac{1}{cos(x^2)}\right)^2= \frac{1}{cos^2(x^2)}$

Are these statements true?

so from all of this we gather;

1. $sec(x)= \frac{1}{cos(x)}$

2. $sec(x^2)= \frac{1}{cos(x^2)}$

3. $sec^2(x)= \left(\frac{1}{cos(x)}\right)^2= \frac{1}{cos^2(x)}$

4. $sec^2(x^2)= \left(\frac{1}{cos(x^2)}\right)^2= \frac{1}{cos^2(x^2)}$

Are these statements true?
Yes, these statements are true.
But #4 isn't related (as far as I can tell) to the original problem.

8. Thanks,

...thats, ok, i need the last one for a similar question, but i didnt feel that i needed to post the whole thing up,