1. ## Trig Identity Help

$
\csc x - \cot x = \frac{{\sin x}}
{{1 + \cos x}}
$

If you want to know where I keep getting stuck, it is at

$
\frac{1}
{{\sin x}} - \frac{{\cos x}}
{{\sin x}} = \sin x + \frac{{\sin x}}
{{\cos x}}
$

2. Originally Posted by Zabaf

$
\csc x - \cot x = \frac{{\sin x}}
{{1 + \cos x}}
$

If you want to know where I keep getting stuck, it is at

$
\frac{1}
{{\sin x}} - \frac{{\cos x}}
{{\sin x}} = \sin x + \frac{{\sin x}}
{{\cos x}}
$
first: if you are proving an identity, you are not allowed to work on both sides at the same time. work on one side at a time, either trying to change one side to look like the other, or changing both sides to the same thing

second: you cannot split the denominator of a fraction!!!!! (i want to put more exclamation marks, but i don't want you to think i am angry with you )

in general $\frac a{b + c}{~\color{red} \ne~} \frac ab + \frac ac$

3. i dont know how to use the forum's fraction option.

where did you got your sinX on the second side?

here is how you do it anyway,
cross multiply the first side(continued from the one that you are working) and you get sinX-sinXcosX/sinX squared, then factor the numerator, you get sinX(1-cosX). cancel the sinX and the sinX from the denominator removing its square. then you get (1-cosX)/sinX on the first side and sinX/1+cosX on the second. then from a/b=c/d equals a.d=c.b we get 1 - cosx squared = sinX squared.

thus we have proven the identity~

4. Originally Posted by Jhevon
second: you cannot split the denominator of a fraction!!!!! (i want to put more exclamation marks, but i don't want you to think i am angry with you )
You can probably get away with infinitely repeating exclamation points by doing this: $\bar{!}$

--Chris

5. why would you cross multiply?
the have a common denimator dont you you just subtract which gives you

$\frac{1-cosx}{sinx}$

and you cross x multipling seems wrong

6. Originally Posted by proview
i dont know how to use the forum's fraction option.

where did you got your sinX on the second side?

here is how you do it anyway,
cross multiply the first side(continued from the one that you are working) and you get sinX-sinXcosX/sinX squared, then factor the numerator, you get sinX(1-cosX). cancel the sinX and the sinX from the denominator removing its square. then you get (1-cosX)/sinX on the first side and sinX/1+cosX on the second. then from a/b=c/d equals a.d=c.b we get 1 - cosx squared = sinX squared.

thus we have proven the identity~
Originally Posted by sk8erboyla2004
why would you cross multiply?
the have a common denimator dont you you just subtract which gives you

$\frac{1-cosx}{sinx}$

and you cross x multipling seems wrong
you cannot cross-multiply or subtract. you are not solving an equation here, you are proving a trig identity.

follow the directions in post #2

7. Originally Posted by Jhevon
you cannot cross-multiply or subtract. you are not solving an equation here, you are proving a trig identity.

follow the directions in post #2
he meant, subtract as in write it as 1 minus cosX. of course you dont subtract it literally(i.e. perform the operation).

why cant cross multiply? im just basically following the indicated operation. anyway, that is how we were thought and it was proven nonetheless. point out exactly which part i did wrong and then we'll talk.

@sk8erboyla2004
both will arrive at the same answer anyway. look at my answer after cancelation.lol any fraction with common denominator, you can do the copy denominator rule or do what i just did. both will be correct just that the copy the denominator rule will be in the lower term.

p.s.
jhevon maybe you can help me with my geometry? im having hard time picturing the figure. http://www.mathhelpforum.com/math-he...ian-plane.html

8. then from a/b=c/d equals a.d=c.b we get 1 - cosx squared = sinX squared.

I understood up untill this part, Any more help?

9. as i said, when proving trig identities, you can't let the sides interact with each other. work on one, or one at a time.

we wish to prove $\csc x - \cot x \equiv \frac {\sin x}{1 + \cos x}$

Consider the LHS:

$\csc x - \cot x = \frac 1{\sin x} - \frac {\cos x}{\sin x}$

$= \frac {1 - \cos x}{\sin x}$

$= \frac {\sin x}{\sin x} \cdot \frac {1 - \cos x}{\sin x}$

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

.
.
.

$= RHS$

can you finish up?

10. Sadly no, I really suck at these trig identities, I just cant get a grip on them.

11. Originally Posted by Zabaf
Sadly no, I really suck at these trig identities, I just cant get a grip on them.
hint: $\sin^2 x = 1 - \cos^2 x$

and recall the formula for the difference of two squares

12. I'm giving up on this problem. I have been working on it for about 6 hours and I cant handle it anymore. What can you suggest that will help you get better at identities, that dont involve me banging my head against a wall?

Irealise that this problem is pretty simple, but it is hard for me

13. Originally Posted by Zabaf
then from a/b=c/d equals a.d=c.b we get 1 - cosx squared = sinX squared.

I understood up untill this part, Any more help?

if you understood until that part then good.

a/b=c/d equals a.d=c.b

its as is.

if the fractions are equal, the numerator of the first fraction times the denominator of the second fraction will always be equal to the denominator of the second fraction times the numerator of the first fraction.

i.e.

1
/2 = 2/4

1 x 4 = 4
2 x 2 = 4

or

3/6 = 6/12

3 x 6 = 36
6 x 6 = 36

colored numbers are numerators.