# Indentites - Calculus (newbie)

1. Apr 21, 2004

### fergie

Hi all.
Hey - i'm new here - so please excuse me if i have posted in the wrong thread etc
Now, for the details.
Firstly, I'm from New Zealand, 16yrs - at High School.
I'm doing NCEA (new system) level 3 Calculus (last year High School Calc)
I came here mainly to look around - but i have a problem, so i thought I might as well ask.

We're doing Trigonometry at the moment - and we're touching on Indentites - and proving them.
Me and my mate have been going through doing the worksheet we got given for the holidays - we've managed them all ok, except from one, so i thought i'd come hear for help (like i said earlier - this may be the wrong thread - o, and what is "college" and Grades K-12 - what would i fit into...)

I'm not to sure if this is standard stuff or not... but anyway, here goes...

We have to prove that:

cos²x cosec²x - sin²x sec²x = 4 cot2x cosec2x

Now, like I said - i tried it by myself - no luck, went to my classmate - we spent about 3 hrs straight, and 20-30 pages, and were completey puzzled.

If I do it in the simplistic way & change the Left side, I change:
cosec²x - which is 1/sin²x (correct?)

and sec²x to 1/cos²x (correct?)
mulitply cos²x with sin²x = cos²x/sin² x
and the other one = sin²x/cos²x
Subtract the two, and i get = 1 -- which is not true...

Another method we tried was...
as cosec²x = cot²x + 1
and
sec²x = tan²x + 1
you can subsitute them in, then change the cot²x to cos²x/sin²x (is that true? will they both be squared?)
and the tan²x to sin²x/cos²x

but even with this way we ended up in a jumble mess

We also tried starting by changing the right hand side (the 4 cot2x cosec2x) - even though we were talk not to use this method UNLESS YOU HAD TO, but we still got no where.

There were heaps of other methods we used - but I'd be here all night typing them up.
So if anyone has any ideas - i really only needed a pointer, or hint - i'll be able to sus out the rest! and I don't like 'cheating'

Thanks
Alex

2. Apr 21, 2004

### recon

I don't know if there are shorter ways of doing it, but the whole solution took me one page, and I have really small handwriting. Anyway, here's where you went wrong:

Check it again. When you subtract the two, it does not result in a one. Correct this and continuing solving it using this method (not the second one).

3. Apr 21, 2004

### recon

It also helps to change the Right-hand Side. In order to change the RHS, it will be useful for you to know the double-angle formulas. Do you know them already?

4. Apr 21, 2004

### HallsofIvy

Staff Emeritus
For most people, it is simplest to convert everything to sine and cosine- since we are more used to them. In this example, cos2(x)cosec2(x) becomes cos2(x)/sin2(x) and sin2(x)sec(x) becomes
sin2(x)/cos2(x). Now get common denominators and subtract the fractions:
$$\frac{cos^2(x)}{sin^2(x)}- \frac{sin^2(x)}{cos^2(x)}= \frac{cos^4(x)- sin^4(x)}{sin^2(x)cos^2(x)}$$

You should be able to recognize cos4(x)- sin4(x) as a "difference of squares": it is equal to (cos2(x)- sin2(x)(cos2(x)+ sin2(x))- and we all know what cos2(x)+ sin2(x) is!

The left-hand side reduces to
$$\frac{cos^2(x)- sin^2(x)}{sin^2(x)cos^2(x)}$$

I seem to recall an identity that says "cos2(x)- sin2(x)= cos(2x)" as well as "2sin(x)cos(x)= sin(2x)". Since the right-hand side has "2x", those should be useful! (Aha! Now I see where that "4" on the right-hand side came from!)

5. Apr 22, 2004

### fergie

Hi.
Hey - thanks guys for that.
I went bak thro all our working - dunno why i thought that equalled one...
I see i did it like that (=1) on the very first attempt - and when i went through with my mate - we only spent a few mins on it (doing it that way) and we got one aswell - our working was correct up to the subtraction part - i guess we just had a block or something.

Anyway - thanks - i thought it was something simple - or something to hard that we hadn't been taught.
I'll go work on it, and get back to you

6. Apr 22, 2004

### fergie

Hmmmm.
Ok, i'm a little confused.
We only really touched on Double Angles on the last 2 days of the term (and you know what the last few days of term are like...)
Yeah, i have all those identities of them - so thats ok.
Going 'HallsofIvy' way - i also get it down to:

cos²(x)-sin²(x)
--------------
sin²(x)cos²(x)

As cos2(x)+ sin2(x) = 1

Thats all cool, but i trip over at the next stage.
Yup, you're right on "cos2(x)- sin2(x)= cos(2x)" - but i also have noted that it can also become =1-2sin²(x)

But what you do at the next stage I don't understand...
>as well as "2sin(x)cos(x)= sin(2x)"

What is that referring to?

I get to:
Cos(2x) or 1-2sin²(x)
----------- -----------
sin²(x)cos²(x) sin²(x)cos²(x)

hmmmm...
nope, I'm lost :(

And 'recon' - i have a few Q's about Double Angles and changing the right hand side:
1) when you change 4cot(2x) - what does it work out to - (what happens to the 2x)
Does it become : 4cos(2x)
--------
sin (2x) or is the 2x only on one side? and if so, which, and why?

So if my above method is correct - i get this:
4cos(2x)
--------
sin²(4x)

And then - changing 4cos(2x) - using the unfamilar Double Angles -
2cos²(x)-2sin²(x)

na, i'm sure i'm just drifting off here - i should have taken more notice in that last class of Calc!

But going backwards - I some how have to get sin²(x)cos²(x) to = 4/sin (2x) then i'm sweet!
But thats the link I'm missing...

7. Apr 22, 2004

### recon

tan(2x) = [1 - tan^2(x)]/2 tan (x)

Since cot(x) = 1/tan(x) so

cot(2x) = 1/tan (2x) = 2 tan (x)/[1 - tan^2(x)]

8. Apr 22, 2004

### recon

I did not use the double angle formula for cos (2x) in solving the problem.

9. Apr 25, 2004

### fergie

Ok, i'm still not thinking on the right lines here....

I can get to this:

cos(2x)
-----------
sin²(x)cos²(x)

on left side
and this:

4/tan(2x) . 1/sing(2x) on right

And changing the Tan like Recon said...

4(1-tan²(x) 1
----------- . ----
2tan x sin(2x)

but thats just going to end up in a messs.

I'm sure I'm missing something on the left side.

Recon - did you get to where I am now? If not - could you tell me which side i have went rong on, and how to fix it up.
And did you alter both sides so they become equal - but not showing what the question asked (i hope thats worded right...!)

10. Apr 25, 2004

### Xishan

once you get cos(2x) in the numerator and sin(2x)/2 squared in the denomenator, the problem is solved!!!

Look how,

cos(2x)/(sin(2x)/2) = 2cot(2x) right!
and 1/(sin(2x)/2) = 2cosec(2x) OK!

now multiply the two and you get 4cot(2x)cosec(2x)

11. Apr 25, 2004

### recon

Fergie, forget about my way of solving this problem and stick with HallsofIvy. His method is easier to understand. Mine goes on for many many lines. Anyway, I thought I might outline HallsofIvy and Xishan's solution more clearly:

$$2sin(x)cos(x) = sin(2x)$$ --> Double angle formula

So $$sin(x)cos(x) = \frac{sin(2x)}{2}$$
And squaring both sides,

$$sin^2(x)cos^2(x) = \frac{sin^2(2x)}{4}$$ - Just plug this in the equation below.

As others have pointed out $$cos^2(x)- sin^2(x)= cos(2x)$$

So after doing some simple algebra you get

$$\frac{4cos(2x)}{sin^2(2x)}$$

Then have a look at Xishan's post.

Last edited: Apr 25, 2004
12. Apr 25, 2004

### Xishan

Thanks for further clarifying my point!

13. Apr 27, 2004

### fergie

hi.
thanks all!
I think i have it sorted.
You guys cleared a few things up for me - thanks a heap.

No doubt i'll be back again in the future
Thanks
Alex