[Help] Verify the identity problem.

[Tickle]

I ran across this problem a few days ago. I've been trying to solve it on my own. With no movement forward.

Problem:

(sinx + cosx)^2 / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2


The problem I come across is:

The trig identity is:
sin^2x + cos^2x = 1 but is it also true that sinx + cosx = 1 ? without the power of 2?

Also, I know that ( sinx + cosx )^2 is not sin^2x + cos^2x.

It is instead:
(sinx + cosx)(sinx + cosx)

Now I usually go ahead and try:

( sinx * sinx ) + ( sinx * cosx ) + ( cosx * sinx ) + ( cosx * cosx ) which yields:

sin^2x + 2sinxcosx + cos^2x

Now I see the identity there. So I take 1:

2sinxcosx / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2 Which is my problem now right?

Not sure what to do next.

I think this problem is just over my head.

 

[Deleted member 4993]

I ran across this problem a few days ago. I've been trying to solve it on my own. With no movement forward.

Problem:

(sinx + cosx)^2 / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2


The problem I come across is:

The trig identity is:
sin^2x + cos^2x = 1 but is it also true that sinx + cosx = 1 ? without the power of 2?

Also, I know that ( sinx + cosx )^2 is not sin^2x + cos^2x.

It is instead:
(sinx + cosx)(sinx + cosx)

Now I usually go ahead and try:

( sinx * sinx ) + ( sinx * cosx ) + ( cosx * sinx ) + ( cosx * cosx ) which yields:

sin^2x + 2sinxcosx + cos^2x

Now I see the identity there. So I take 1:

2sinxcosx / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2 Which is my problem now right?

Not sure what to do next.

I think this problem is just over my head.

First thing you need to do is post the problem (and your work) with proper parentheses so that we can follow the hierarchy of operations. As posted, your problem looks like:

$\displaystyle \displaystyle{\frac{[sin(x) + cos(x)]^2}{sin^2(x)} - cos^2(x) \ = sin^2(x) - \frac{cos^2(x)}{[sin(x) - cos(x)]^2}}$

That is not an identity that can be proved - so you need to fix it!!

 

[Tickle]

Original Problem

(sinx + cosx)^2 / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2


another way to put it


[(sinx + cosx)^2] / [sin^2x - cos^2x] = [sin^2x - cos^2x] / [( sinx - cosx )^2]


another way to put it

(sinx + cosx)^2 over sin^2x - cos^2x is equal to sin^2x - cos^2x over ( sinx - cosx )^2


There is 3 ways. I'm sure you can get the problem now?

 

[Ishuda]

Original Problem

(sinx + cosx)^2 / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2


another way to put it


[(sinx + cosx)^2] / [sin^2x - cos^2x] = [sin^2x - cos^2x] / [( sinx - cosx )^2]


another way to put it

(sinx + cosx)^2 over sin^2x - cos^2x is equal to sin^2x - cos^2x over ( sinx - cosx )^2


There is 3 ways. I'm sure you can get the problem now?

No, not really. Since you have used the grouping symbols, you appear to know how they are used and should be intelligent enough to know that those three ways are not equivalent under the general rules of mathematics. So would you please answer the original question?

BTW: Have you read the "rules" about posting?
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting
and the posting guide lines referenced there
http://www.freemathhelp.com/forum/threads/41536-Read-Before-Posting!!
I quote from the latter in part
"Be nice. Everyone here is a volunteer, so treat them with respect and we will (probably) be nice in return."

 

[Steven G]

Original Problem

(sinx + cosx)^2 / sin^2x - cos^2x = sin^2x - cos^2x / ( sinx - cosx )^2


another way to put it


[(sinx + cosx)^2] / [sin^2x - cos^2x] = [sin^2x - cos^2x] / [( sinx - cosx )^2]


another way to put it

(sinx + cosx)^2 over sin^2x - cos^2x is equal to sin^2x - cos^2x over ( sinx - cosx )^2


There is 3 ways. I'm sure you can get the problem now?

So there are 3 identity problems.