# Thread: Trig Expressions and Identities.

1. ## 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

2. Originally Posted by amanda0603
1. Mulitply and simplify-
cosXsinX(secX-cscX)
I got 1 by turning it into 1 but Im not 100% sure.
I'm sorry, but I don't understand what this means...? I mean, yes, you can "turn it into 1" and thus "get" 1, but what was your mathematical reasoning for "turning" something into some value you picked?

A good way to start is to convert everything into sines and cosines:

. . . . .$\displaystyle \left(\cos(x)\sin(x)\right)\left(\frac{1}{\cos(x)} \, -\, \frac{1}{\sin(x)}\right)$

. . . . .$\displaystyle \left(\frac{\cos(x)\sin(x)}{1}\right)\left(\frac{\ sin(x)\, -\, \cos(x)}{\cos(x)\sin(x)}\right)$

How did you go from the above to "1"?

Originally Posted by amanda0603
2. Simplify expression-
(15tanXcscX-3cscX)/(9tanXcscX-3cscX)
Factor, and then see what you can do with whatever is left:

. . . . .$\displaystyle \frac{3\csc(x)\left(5\tan(x)\, -\, 1\right)}{3\csc(x)\left(3\tan(x)\, -\, 1\right)}$

. . . . .$\displaystyle \frac{5\tan(x)\, -\, 1}{3\tan(x)\, -\, 1}$

I don't know how "simplified" your book wants you to go, so you'll have to decide where to go from the above.

Originally Posted by amanda0603
3. Establish the Indentity
(1-sin2X)/(sinX-cscX)=-sinX
The customary technique is to work on the more-complicated side, converting everything to sines and cosines, usually, and applying whatever identities seem useful.

. . . . .$\displaystyle \frac{1\, -\, \sin(2x)}{\sin(x)\, -\, \frac{1}{\sin(x)}}$

. . . . .$\displaystyle \frac{\left(1\, -\, 2\sin(x)\cos(x)\right)}{\left(\frac{\sin^2(x)\, -\, 1}{\sin(x)}\right)}$

. . . . .$\displaystyle \left(\frac{1\, -\, 2\sin(x)\cos(x)}{1}\right)\left(-\frac{\sin(x)}{\cos^2(x)}\right)$

...and so forth.

Originally Posted by amanda0603
4. Establish the identity
cotX(cotX+tanX)=csc^2X
The left-hand side would appear to be the more-complicated one, so:

. . . . .$\displaystyle \left(\frac{\cos(x)}{\sin(x)}\right)\left(\frac{\c os(x)}{\sin(x)}\, +\, \frac{\sin(x)}{\cos(x)}\right)$

. . . . .$\displaystyle \left(\frac{\cos(x)}{\sin(x)}\right)\left(\frac{\c os^2(x)\, +\, \sin^2(x)}{\sin(x)\cos(x)}\right)$

. . . . .$\displaystyle \left(\frac{\cos(x)}{\sin(x)}\right)\left(\frac{1} {\sin(x)\cos(x)}\right)$

...and so forth.

Originally Posted by amanda0603
5. Establish the indentity
1- (cos^2X/1+sinX)=sinX
Again, the left-hand side appears to be more complex, so:

. . . . .$\displaystyle 1\, -\, \frac{1\, -\, \sin^2(x)}{1\, +\, \sin(x)}$

. . . . .$\displaystyle 1\, -\, \frac{(1\, -\, \sin(x))(1\, +\, \sin(x))}{1\, +\, \sin(x)}$

...and so forth.

Originally Posted by amanda0603
6. (sinX)/(1+cosX)+(1+cosX)/(sinX)= 2cscX
What were this instructions on this one? What have you tried? How far have you gotten? Where are you stuck?

Originally Posted by amanda0603
I've been working on this forever
In future, it might help if you showed that work, so that we can "see" where you're having trouble and then provide intelligent advice. Thank you!