# Trig Expressions and Identities.

• Mar 22nd 2009, 06:43 PM
amanda0603
Trig Expressions and Identities.

1. Mulitply and simplify-
cosXsinX(secX-cscX)
I got 1 by turning it into 1 but Im not 100% sure.

2. Simplify expression-
(15tanXcscX-3cscX)/(9tanXcscX-3cscX)

3. Establish the Indentity
(1-sin2X)/(sinX-cscX)=-sinX

4. Establish the identity
cotX(cotX+tanX)=csc^2X

5. Establish the indentity
1- (cos^2X/1+sinX)=sinX

6. (sinX)/(1+cosX)+(1+cosX)/(sinX)= 2cscX

I've been working on this forever and cant seem to get these answers, any help is awesome! THANKS SO MUCH
• Mar 22nd 2009, 07:00 PM
coolguy99
For the first one, I'm not sure how you got 1 hehe

cos(x)sin(x) (sec(x)-csc(x))

(cos(x)sin(x)/cos(x)) - (cos(x)sin(x)/sin(x))

sin(x) - cos(x)

and for the fourth one..

Cot(x) (cot(x)+tan(x)) = csc^2(x)

multiply out..

cot^2(x) + (cot(x)tan(x)) = csc^2(x)

cot^2(x) + 1 = csc^2(x)

which is a common trig identity

and for the last one...

make a common denominator, and remember that sin^2(x) + cos^2(x) = 1. then just do some algebra magic :)
• Mar 22nd 2009, 07:14 PM
e^(i*pi)
Quote:

Originally Posted by amanda0603

1. Mulitply and simplify-
cosXsinX(secX-cscX)
I got 1 by turning it into 1 but Im not 100% sure.

$sec(x)-csc(x) = \frac{1}{cos(x)} - \frac{1}{sin(x)}$. Get the same denominator which will then cancel with the $cos(x)sin(x)$ also present

$\frac{1}{cos(x)} - \frac{1}{sin(x)} = \frac{sin(x)-cos(x)}{cos(x)sin(x)}$ so the final answer is

$\frac{cos(x)sin(x)(sin(x)-cos(x)}{cos(x)sin(x)} = sin(x)-cos(x)$

Quote:

2. Simplify expression-
(15tanXcscX-3cscX)/(9tanXcscX-3cscX)
Factor and cancel:
$\frac{3csc(x)(5tan(x)-1)}{3csc(x)(3tan(x)-1)} = \frac{5tan(x)-1}{3csc(x)-1}$

Quote:

3. Establish the Indentity
(1-sin2X)/(sinX-cscX)=-sinX
Quote:

4. Establish the identity
cotX(cotX+tanX)=csc^2X
$cot(x)+tan(x) = \frac{1+tan^2(x)}{tan(x)} = \frac{sec^2(x)}{tan(x)}$

$\frac{cot(x)sec^2(x)}{tan(x)} = \frac{cos^2(x)sec^2(x)}{sin^2(x)} = csc^2(x)$

Quote:

5. Establish the indentity
1- (cos^2X/1+sinX)=sinX
$\frac{cos^2(x)}{1+sin(x)} = \frac{1-sin^2(x)}{1+sin(x)}$

use the difference of two squares on the numerator and 1+sin(x) should cancel