# Math help part 2!



## musiclover55 (Feb 7, 2011)

I suck at this. It's Analytic Trigonometry if that helps. Unit circle, all that jazz.
And if it's not too much trouble can you explain the steps of solving? All the help is appreciated!

*1. Simplify: cscx/(over) tanx+cotx*

*2. Perform the addition and simplify: tanx/cscx + sinx/tanx*

*7. Find all solutions in the interval [0,2(pi)]: 2cos^2 (2)-1=0

14. Find all solutions in the interval [0,2(pi)]: 2sin^2 (x/4) -3cos(x/4)=0

16. Perform the addition and simplify: 1/1+sinx + 1/1-sinx*


----------



## uffie (May 11, 2010)

show me your steps and ill show you where you went wrong. No one should help you without you atleast trying it.


----------



## musiclover55 (Feb 7, 2011)

uffie said:


> show me your steps and ill show you where you went wrong. No one should help you without you atleast trying it.


Ok then.

1. I used the reciprocal identity and changed cscx to 1/sinx and to change cot x to 1/tanx. Then I crossed out both tanx's and got 1/sinx OVER 1.... which was not an answer choice.

2. changed sinx to 1/cscx and crossed out both cscx's. Was left with tanx/tanx equals 1... which wasn't right.

7. I tried to multiply 2cos^2x by 2x.

14. I tried to type it into my Casio Fx-115ES and got "Can't Solve".

16. I just straight up didn't know what to do.


----------



## Ape in space (May 31, 2010)

musiclover55 said:


> Ok then.
> 
> 1. I used the reciprocal identity and changed cscx to 1/sinx and to change cot x to 1/tanx. Then I crossed out both tanx's and got 1/sinx OVER 1.... which was not an answer choice.
> 
> 2. changed sinx to 1/cscx and crossed out both cscx's. Was left with tanx/tanx equals 1... which wasn't right.


You can't cancel out the tan(x) in #1 or the csc(x) in #2. You can only cancel things out if you have an equation and you have the same factor on both sides of the equation. Here you're just trying to simplify an expression so you have to keep the tan(x) in #1 and csc(x) in #2. Also, in #2, you have confused addition with multiplication: you have tan(x) + 1/tan(x), which is different from tan(x) ( 1 / tan(x) ), so it doesn't equal 1.

I won't give the answers, but I'll tell you how to start.

1,2. If you are having trouble with these expressions, the best thing to do is to write everything in the expressions in terms of only sines and cosines. Remember that tan(x) = sin(x) / cos(x), that csc(x) = 1/sin(x), and cot(x) = 1/tan(x) = cos(x) / sin(x). Then simplify the expression and see what you get. I've done the first couple of steps in the attached file to show you how to start.

7, 14. Think about how you would solve a problem that looked like this: cos(x) = 1/3 (I'm just making up those numbers). If you know how to solve that, then try to simplify your problem so that it looks like this type of problem, which you can solve.

16. You have two fractions to add. Use the usual procedure for adding two fractions, and then simplify both the numerator and denominator, keeping in mind the standard trig identities, and see what you get.


----------



## musiclover55 (Feb 7, 2011)

Ape in space said:


> You can't cancel out the tan(x) in #1 or the csc(x) in #2. You can only cancel things out if you have an equation and you have the same factor on both sides of the equation. Here you're just trying to simplify an expression so you have to keep the tan(x) in #1 and csc(x) in #2. Also, in #2, you have confused addition with multiplication: you have tan(x) + 1/tan(x), which is different from tan(x) ( 1 / tan(x) ), so it doesn't equal 1....


Ok for :
#16 I got (1-sinx)+(1+sinx) / (1+sinx)(1-sinx)... so do I just cancel them all out and get 0? The remaining answer choices were 0, 2sec^2 x, and 2cos^2 x.

#1 (from where you left off) for the bottom fractions I combined the fractions and the numerator and denominator = 1 so I ended up with 1 * 1/sinx and switched it back to cscx. The answers for that are cos x + tan x, sin^2 + cosx, or cos x.


----------



## SuperSky (Feb 16, 2011)

musiclover55 said:


> Ok for :
> #16 I got (1-sinx)*+*(1+sinx) / (1+sinx)(1-sinx)... so do I just cancel them all out and get 0? The remaining answer choices were 0, 2sec^2 x, and 2cos^2 x.


Nope, expand out the brackets and you'll find that they don't cancel out, because the terms in the numerator are added, whereas they're multiplied in the denominator.
So expand, see what you get, and see if there's any trig identity you can use.



musiclover55 said:


> #1 (from where you left off) for the bottom fractions I combined the fractions and the numerator and denominator = 1 so I ended up with 1 * 1/sinx and switched it back to cscx. The answers for that are cos x + tan x, sin^2 + cosx, or cos x.


Might want to try adding the fractions again. It shouldn't give 1/1 for that addition.


----------



## Ape in space (May 31, 2010)

musiclover55 said:


> #1 (from where you left off) for the bottom fractions I combined the fractions and the numerator and denominator = 1


You should get 1 for the numerator, but not the denominator. When you add two fractions, the numerator comes from cross-multiplying and adding, while for the denominator you multiply the two denominators, like this:

a/c + b/d = (ad + bc) / (cd)

Try this procedure for the bottom fractions and see what you get after you simplify.



> #16 I got (1-sinx)+(1+sinx) / (1+sinx)(1-sinx)... so do I just cancel them all out and get 0? The remaining answer choices were 0, 2sec^2 x, and 2cos^2 x.


You have to add the two terms in the numerator, and then multiply the two things in the denominator using the usual procedure for multiplying terms of this type. To get the final answer, you will need to make use of one of the standard trigonometric identities.


----------



## musiclover55 (Feb 7, 2011)

Hmm ok. 
Thank you all very much for the help! 
I'm just done with this.
I'm a little better because of the clarifications so... thanks again!!


----------

