Indentites - Calculus (newbie)

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In summary, Alex has difficulty solving a problem involving proving that: cos²x cosec²x - sin²x sec²x = 4 cot2x cosec2x. He goes to his classmates for help, but they are also unable to solve the problem. They try multiple methods before finding one that Alex can understand.
  • #1
fergie
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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
 
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  • #2
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:

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...

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
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
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:
[tex]\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)}[/tex]

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
[tex]\frac{cos^2(x)- sin^2(x)}{sin^2(x)cos^2(x)}[/tex]

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
Hi.
Hey - thanks guys for that.
I went bak thro all our working - don't know 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 as well - 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
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 that's 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 that's the link I'm missing...
 
  • #7
fergie said:
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)

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
HallsofIvy said:
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!)

I did not use the double angle formula for cos (2x) in solving the problem.
 
  • #9
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 that's 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 that's worded right...!)
 
  • #10
'HallsofIvy' has already solved your problem Sir

HallsofIvy said:
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:
[tex]\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)}[/tex]

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
[tex]\frac{cos^2(x)- sin^2(x)}{sin^2(x)cos^2(x)}[/tex]

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!)

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
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:

[tex]2sin(x)cos(x) = sin(2x)[/tex] --> Double angle formula

So [tex]sin(x)cos(x) = \frac{sin(2x)}{2}[/tex]
And squaring both sides,

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

HallsofIvy said:
The left-hand side reduces to
[tex]\frac{cos^2(x)- sin^2(x)}{sin^2(x)cos^2(x)}[/tex]

As others have pointed out [tex]cos^2(x)- sin^2(x)= cos(2x)[/tex]

So after doing some simple algebra you get

[tex]\frac{4cos(2x)}{sin^2(2x)}[/tex]

Then have a look at Xishan's post.
 
Last edited:
  • #12
Thanks for further clarifying my point!
 
  • #13
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
 

1. What is an identity in calculus?

An identity in calculus is a mathematical statement that is always true, regardless of the values of the variables involved. It can be thought of as an equation that is always balanced, no matter what values are substituted in for the variables.

2. How do I use identities in calculus?

Identities in calculus are used to simplify expressions and equations. By using identities, you can transform complex expressions into simpler ones, making them easier to work with and solve.

3. Can I create my own identities in calculus?

Yes, you can create your own identities in calculus by manipulating existing equations and expressions. However, it is important to make sure that your identity is true for all values of the variables involved.

4. What is the difference between an identity and an equation?

An identity is a statement that is always true, while an equation is a statement that is only true for specific values of the variables involved. In other words, an identity is a more general form of an equation.

5. Are identities only used in calculus?

No, identities are used in various branches of mathematics, including algebra, trigonometry, and calculus. They are a fundamental concept in mathematics and have many applications in problem-solving and proof-writing.

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