# proving expressions.

• Jul 31st 2009, 12:29 PM
Tweety
proving expressions.
Quote:

Given that $\displaystyle a = 4 sec x , b = cosx , c = cotx$

a) express b in terms of a
b) show that $\displaystyle c^{2} = \frac{16}{a^{2} - 16}$

a) $\displaystyle b = cosx$
$\displaystyle a = \frac{4}{cosx}$
$\displaystyle a = \frac{4}{b}$
$\displaystyle b = \frac{4}{a}$

having trouble with part 'b';
• Jul 31st 2009, 12:33 PM
skeeter
Quote:

Originally Posted by Tweety
a) $\displaystyle b = cosx$
$\displaystyle a = \frac{4}{cosx}$
$\displaystyle a = \frac{4}{b}$
$\displaystyle b = \frac{4}{a}$

having trouble with part 'b';

$\displaystyle \frac{16}{(4\sec{x})^2 - 16}$

$\displaystyle \frac{16}{16\sec^2{x} - 16}$

$\displaystyle \frac{1}{\sec^2{x} - 1}$

$\displaystyle \frac{1}{\tan^2{x}}$

$\displaystyle \cot^2{x} = c^2$
• Jul 31st 2009, 12:54 PM
Tweety
Quote:

Originally Posted by skeeter
$\displaystyle \frac{16}{(4\sec{x})^2 - 16}$

$\displaystyle \frac{16}{16\sec^2{x} - 16}$

$\displaystyle \frac{1}{\sec^2{x} - 1}$

$\displaystyle \frac{1}{\tan^2{x}}$

$\displaystyle \cot^2{x} = c^2$

I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16}$? Why are you doing $\displaystyle c^{2} = cot^2{x}$?

This is what I have done but it does not fall together!

$\displaystyle c = cotx , c^{2} = cot^{2}x$

$\displaystyle b = \frac{4}{a} , b^{2} = \frac{16}{a^{2}}$

$\displaystyle cos^{2}x = \frac{16}{a^{2}}$

$\displaystyle 1-sin^{2}x = \frac{16}{a^{2}}$

$\displaystyle a^{2}(1-sin^{2}x) = 16$

$\displaystyle a^{2} = \frac{16}{1-sin^{2}x}$

$\displaystyle a^{2} = \frac{16}{1} - \frac{16}{sin^{2}x}$

$\displaystyle a^{2} = 16 - 16cosec^2x$

$\displaystyle a^2 = 16 - 16(1+cot^{2}x)$

$\displaystyle a^{2}= 16- 16(1+c^{2})$

$\displaystyle a^{2} -16 =-16( 1+c^{2})$

is this right so far? But I am stuck here,
• Jul 31st 2009, 01:00 PM
skeeter
Quote:

Originally Posted by Tweety
I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16}$? Why are you doing $\displaystyle c^{2} = cot^2{x}$?

This is what I have done but it does not fall together!

$\displaystyle c = cotx , c^{2} = cot^{2}x$

$\displaystyle b = \frac{4}{a} , b^{2} = \frac{16}{a^{2}}$

$\displaystyle cos^{2}x = \frac{16}{a^{2}}$

$\displaystyle 1-sin^{2}x = \frac{16}{a^{2}}$

$\displaystyle a^{2}(1-sin^{2}x) = 16$

$\displaystyle a^{2} = \frac{16}{1-sin^{2}x}$

$\displaystyle a^{2} = \frac{16}{1} - \frac{16}{sin^{2}x}$ this step is incorrect ... you cannot "split" the denominator like that.

...
• Jul 31st 2009, 01:07 PM
Tweety
Quote:

Originally Posted by skeeter
...

Okay, so how would I 'carry' on from there?

Thanks
• Jul 31st 2009, 01:34 PM
AlephZero
Quote:

Originally Posted by Tweety
Okay, so how would I 'carry' on from there?

Thanks

$\displaystyle \sin^2 x + \cos^2 x =1,$ so $\displaystyle 1-\sin^2 x =$...
• Jul 31st 2009, 02:17 PM
skeeter
Quote:

Originally Posted by Tweety
I dont really understand what you did, aren't you meant to prove that $\displaystyle c^{2} = \frac{16}{a^{2}-16}$? Why are you doing $\displaystyle c^{2} = cot^2{x}$?

I started with $\displaystyle \frac{16}{a^2-16}$ and showed it was equal to $\displaystyle c^2$.

note that one may start on either side of an equation and work to get the other side.