Trigonometric Identities

• Aug 23rd 2008, 01:47 PM
magentarita
Trigonometric Identities
Establish each identity below.

(1) 9 sec^2 (x) - 5 tan^2 (x) = 5 + 4 sec^2 (x)

(2) cos(x)/[1 + sin(x)] + [1 + sin(x)]/cos(x) = 2 sec (x)
• Aug 23rd 2008, 01:53 PM
o_O
#1:
Note that $\displaystyle \tan^{2} x = \sec^{2} x - 1$. So substitute it in and simplify.

#2:
Combine into a single fraction:
$\displaystyle \frac{\cos x}{1+\sin x} {\color{blue}\cdot \frac{\cos x}{\cos x} } \: + \: \frac{1+ \sin x}{\cos x} {\color{blue} \cdot \frac{1+ \sin x}{1 + \sin x} }$

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

Expand the numerator and simplify.

Post back if you have any troubles!
• Aug 23rd 2008, 03:26 PM
magentarita
Where???
Quote:

Originally Posted by o_O
#1:
Note that $\displaystyle \tan^{2} x = \sec^{2} x - 1$. So substitute it in and simplify.

#2:
Combine into a single fraction:
$\displaystyle \frac{\cos x}{1+\sin x} {\color{blue}\cdot \frac{\cos x}{\cos x} } \: + \: \frac{1+ \sin x}{\cos x} {\color{blue} \cdot \frac{1+ \sin x}{1 + \sin x} }$

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

Expand the numerator and simplify.

Post back if you have any troubles!

What in the question indicated that you needed to make cosx/cosx
on the left side and 1 + sinx/1 + sinx on the right side?
• Aug 23rd 2008, 04:08 PM
skeeter
Quote:

What in the question indicated that you needed to make cosx/cosx
on the left side and 1 + sinx/1 + sinx on the right side?
basic addition of fractions ... you need the common denominator, $\displaystyle \cos{x}(1 + \sin{x})$, to add the two fractions.
• Aug 23rd 2008, 05:51 PM
magentarita
I see..
Quote:

Originally Posted by skeeter
basic addition of fractions ... you need the common denominator, $\displaystyle \cos{x}(1 + \sin{x})$, to add the two fractions.

It's searching for the LCD, right?
• Aug 23rd 2008, 06:52 PM
o_O
An analogy:

$\displaystyle \frac{1}{2} + \frac{2}{3}$

$\displaystyle = \frac{1}{2} {\color{blue}\cdot \frac{3}{3} } + \frac{2}{3} {\color{blue}\cdot \frac{2}{2}}$

$\displaystyle = \frac{3}{6} + \frac{4}{6}$

$\displaystyle =\frac{7}{6}$
• Aug 24th 2008, 02:48 PM
magentarita
Great...
Great analogy!

Thanks