Mathematics Corner :)

GreyKode said:
read my previous post, I think there seems to be a mistake in the given equation required to prove.
Click to expand...

Yah, I read then edited my post.

The equation that now requires proving is this:

cosec^4 x - cot^4 x = cosec^2 x + cot^2 x
 
solution
cosec^4 x - cot^4 x = cosec^2 x + cot^2 x
(cosec^2 x + cot^2 x)(cosec^2 x - cot^2 x) = cosec^2 x + cot^2 x

cancelling the terms (cosec^2 x + cot^2 x)

you get cosec^2 x - cot^2 x = 1

which was what you proved in a)
 
I see how you're doing it... But I was told to do it kinda like this:

LHS = cosec^4 x - cot^4 x
= (cosec^2 x + cot^2 x)(cosec^2 x - cot^2 x)
.
.
.
= (cosec^2 x + cot^2 x)
= RHS

... You know?
 
okay then

LHS = cosec^4 x - cot^4 x
= (cosec^2 x + cot^2 x)(cosec^2 x - cot^2 x)

but

(cosec^2 x - cot^2 x) = (1-cos^2x)/sin^2x = 1

therefore

LHS = (cosec^2 x + cot^2 x)
=RHS
 
This is why I hate trig identities.

This is the next question following the one I've posted above:

c) Solve, for 90 < x < 180,

cosec^4 x - cot^4 x = 2 - cot x

I started with:

cosec^2 x + cot^2 x = 2 - cot x
1/(sin^2 x) + (cos^2 x)/(sin^2 x) = 2 - cos x/sin x

Now...

If I multiply through by sin x, I get:

1/sin x + cos^2 x/sin x = 2 - cos x

... ?

If I multiply though by sin^2 x, i get:

1 + cos^2 x = 2 - cosx.sinx

... ?
 
^Anyone?

---

I understand why sin2A = 2sinAcosA [well, it's just a rule for me to learn]

But I don't understand how sin5A = 2sin3Acos2A
Why couldn't it be: 2sin2Acos3A? Or does it not make a difference?
 
The Ruler said:
Those are the bits I'm having difficulty understanding. Are those just formulae I'm supposed to know?
Click to expand...

Sorry for the delay, lol....for the first bold part, I was just rearranging tanx = sinx/cosx --> multiplying both sides by cosx gives cosxtanx = sinx.
The second part, yes you need to know the formula - it's just the double angle one: 2sinAcosA = sin2A, only in this case, A = 2k. So:
2sin2kcos2k = sin4k

The Ruler said:
This is why I hate trig identities.

This is the next question following the one I've posted above:

c) Solve, for 90 < x < 180,

cosec^4 x - cot^4 x = 2 - cot x

I started with:

cosec^2 x + cot^2 x = 2 - cot x
1/(sin^2 x) + (cos^2 x)/(sin^2 x) = 2 - cos x/sin x

Now...

If I multiply through by sin x, I get:

1/sin x + cos^2 x/sin x = 2 - cos x

... ?

If I multiply though by sin^2 x, i get:

1 + cos^2 x = 2 - cosx.sinx

... ?
Click to expand...

Did you accidentally go from cosec^4 x to cosec^2 x ??
I would set it out the following way...the main identity you need is cot^2 x + 1 = cosec^2 x, which means
cosec^4 x = (cosec^2 x)^2 = (cot^2 x + 1)^2
So.
cosec^4 x - cot^4 x = 2 - cot x
(cot^2 x + 1)^2 - cot^4 x = 2 - cot x

Multiply out the brackets and simplify:
cot^4 x + 2cot^2 x + 1 - cot^4 x = 2 - cot x
2cot^2 x + 1 = 2 - cot x
2cot^2 x + cotx - 1 = 0
(2cotx + 1)(cotx - 1) = 0......(factorizing)

cotx = -1/2 or cotx = 1
tanx = -2 or tanx = 1......(because tanx = 1/cotx)
I think the only solution for these that's within the given interval is 116.6 degrees...
 
The Ruler said:
^Anyone?

---

I understand why sin2A = 2sinAcosA [well, it's just a rule for me to learn]

But I don't understand how sin5A = 2sin3Acos2A
Why couldn't it be: 2sin2Acos3A? Or does it not make a difference?
Click to expand...

It isn't either, lol...it's sin2Acos3A + cos2Asin3A...there aren't any common terms; you can't simplify further...

Although, of course, you could also divide it up into sinAcos4A + cosAsin4A if you so wished, lol.
 
I'll check your solution to the question a bit later... Thanks.

But:

It isn't either, lol...it's sin2Acos3A + cos2Asin3A...there aren't any common terms; you can't simplify further...
Click to expand...

Heh? I copied that from solutions at the back of my booklet of past papers. The question was:

Find, in terms of pi, the solutions of the equation

sin 5x + sin x = 0

at the back of the booklet, the solution starts with

2 sin 3x cos 2x = 0

Then continues to say either sin 3x = 0 or con 2x = 0 (which is fine... But I don't understand how they got to 2 sin 3x cos 2x from sin 5x + sin x)
 
Okayyy, I had to cheat and look this up :-[; I'd never have thought of this myself, lol...but anyway, you know that:
sin(A+B) = sinAcosB + cosAsinB; and
sin(A-B) = sinAcosB - cosAsinB

Adding the two together, you get:
sin(A+B) + sin(A-B) = (sinAcosB + cosAsinB) + (sinAcosB - cosAsinB)
= 2sinAcosB......(you can see the second terms cancel out)

Apparently, you have to realize that sin5x + sinx can be written as sin(3x+2x) + sin(3x-2x), lol, in which case:
sin5x + sinx = sin(3x+2x) + sin(3x-2x)
= (sin3xcos2x +cos3xsin2x) + (sin3xcos2x - cos3xsin2x)
= 2sin3xcos2x

:mmokay:What an evil question...
 
~Raynn~ said:
Okayyy, I had to cheat and look this up :-[; I'd never have thought of this myself, lol...but anyway, you know that:
sin(A+B) = sinAcosB + cosAsinB; and
sin(A-B) = sinAcosB - cosAsinB

Adding the two together, you get:
sin(A+B) + sin(A-B) = (sinAcosB + cosAsinB) + (sinAcosB - cosAsinB)
= 2sinAcosB......(you can see the second terms cancel out)
Click to expand...

Mmm hmm.

Apparently, you have to realize that sin5x + sinx can be written as sin(3x+2x) + sin(3x-2x), lol, in which case:
sin5x + sinx = sin(3x+2x) + sin(3x-2x)
= (sin3xcos2x +cos3xsin2x) + (sin3xcos2x - cos3xsin2x)
= 2sin3xcos2x
Click to expand...

I understand that sin 5x = sin (3x + 2x)... But I don't understand where the sin x disappeared to... Or how it was included (if it was) in 2sin3xcos2x.
If you can't explain, it's okay. I'll just wait a week and ask the teachers. :)

:mmokay:What an evil question...
Click to expand...

Well, it is from the Solomon Press papers. Ever tried them? They're apparently much harder than questions in actual papers. Or so I was told by the teachers themselves. So if I can do them, I should be able to do the "easier" ones, you know?
 
The Ruler said:
I understand that sin 5x = sin (3x + 2x)... But I don't understand where the sin x disappeared to... Or how it was included (if it was) in 2sin3xcos2x.
Click to expand...
sin5x + sinx = sin(3x+2x) + sin(3x-2x)

Colour-coded, :D ...the sinx is written as sin(3x-2x)... (3x - 2x = x)

If you can't explain, it's okay. I'll just wait a week and ask the teachers. :)
Click to expand...
Lol, good idea...I stink at explaining stuff; my friends complain that I skip half the steps :-\

Well, it is from the Solomon Press papers. Ever tried them? They're apparently much harder than questions in actual papers. Or so I was told by the teachers themselves. So if I can do them, I should be able to do the "easier" ones, you know?
Click to expand...

No, I haven't...mm hmm, that's definitely a good idea, but it makes me feel better, knowing I'm unlikely to get that in an OCR paper, lol...
 
Lol, right...I'm with OCR for every subject, though I remember I did Edexcel gcse maths, lol...

You're doing edexcel? That'd make sense...all the stuff with solving equations using trigonometric identities was in Core 3, not C4, for us...the only trig we have left is in integration :D.
 
I finally got the hang of all C3. Only bits that confuse me a little are graph sketches and domain/range. And a small bit of iteration thingamajig.

So, I thought I'd give C4 a try today... I opened the paper, read the first question and slammed the paper shut (if such a thing is possible). *sigh*
Perhaps tomorrow... Yes, tomorrow seems like a good day to start something new and unfamiliar.
 
Achh, I hated the iteration stuff (I can now say so cheerfully, since C3 is forever behind me :statisfie)

And, :giggling: I so know what you mean...wish you the best of luck with conquering the new and unfamiliar, lol...
 
Iteration is relatively easy.
.
.
.
.
.
.
Until the results that you get don't show the change in sign. That drives me mad because I (still) can't figure out what I'm doing wrong.
 

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