The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Year 13 Maths Help Thread

Scroll to see replies

Original post by RDKGames
1. Get everything under same denominator
2. Tidy up the numerator
3. Cancel any common factors

ImageUploadedByStudent Room1469984130.963724.jpg
Posted from TSR Mobile


ok now im more confused. Wouldnt it be easier to multiply the first fraction by x+2 / x+2
Original post by kiiten
ok now im more confused. Wouldnt it be easier to multiply the first fraction by x+2 / x+2


Sure, but then you would need to multiply it by (x+1)/(x+1) because (x+2)(x+1) is the quadratic on the second fraction so you would only be multiplying by one of the factors when you need both.

All I'm really doing is applying ab+cd=adbd+cbdb=ad+cbbd\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{cb}{db}=\frac{ad+cb}{bd} and simplifying everything. Doesn't get any more complex than that.
(edited 7 years ago)
Original post by RDKGames
Sure, but then you would need to multiply it by (x+1)/(x+1) because (x+2)(x+1) is the quadratic on the second fraction so you would only be multiplying by one of the factors when you need both.

All I'm really doing is applying ab+cd=adbd+cbdb=ad+cbbd\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{cb}{db}=\frac{ad+cb}{bd} and simplifying everything. Doesn't get any more complex than that.


But the first fraction already has x+1 as a denominator??
Original post by kiiten
But the first fraction already has x+1 as a denominator??


Ah right I see what you mean. Yes you can just ignore multiplying by (x+1) then. Though you should still follow that method if the denominator don't share anything in common.
I think my mind is going blank; how do I get the expansion for cosec(2x)cosec(2x)? Do I just use binomial with (1+[sin(2x)−1])−1(1+[sin(2x)-1])^{-1}? Seems a bit long winded. I found the expansion for sin(2x)sin(2x) but unsure how I can use that here.
(edited 7 years ago)
Original post by RDKGames
Ah right I see what you mean. Yes you can just ignore multiplying by (x+1) then. Though you should still follow that method if the denominator don't share anything in common.


Yep. Thanks for your help :smile:
Original post by kiiten
But the first fraction already has x+1 as a denominator??


Yep, you've got the right idea, your method is much more suitable and shows understanding. :-)
Original post by RDKGames
I think my mind is going blank; how do I get the expansion for cosec(2x)cosec(2x)? Do I just use binomial with (1+[sin(2x)−1])−1(1+[sin(2x)-1])^{-1}? Seems a bit long winded. I found the expansion for sin(2x)sin(2x) but unsure how I can use that here.


How many terms do you need? Just differentiate and plug in x=0x=0 into the derivative, same for the second derivative, etc... up till how many terms you need. If you want the general formula, express it in terms of complex exponentials and use their respective taylor expands, you'll get something in the form of Bernoulli numbers.
Original post by Zacken
How many terms do you need? Just differentiate and plug in x=0x=0 into the derivative, same for the second derivative, etc... up till how many terms you need. If you want the general formula, express it in terms of complex exponentials and use their respective taylor expands, you'll get something in the form of Bernoulli numbers.


I just need the first 3, is there no direct way to use the first 3 terms from expansion for sin(2x) for this?
Original post by RDKGames
I just need the first 3, is there no direct way to use the first 3 terms from expansion for sin(2x) for this?


Heuristically, nope. Not worth it.
Original post by Zacken
Heuristically, nope. Not worth it.


Ah right.

Original post by Zacken
Just differentiate and plug in x=0x=0 into the derivative, same for the second derivative, etc... up till how many terms you need.


Do you mean carry out this? cosec(2x) doesn't exist at x=0 :/
asdasd.PNG
(edited 7 years ago)
Original post by RDKGames
Do you mean carry out this? cosec(2x) doesn't exist at x=0 :/
asdasd.PNG


Can you post a picture of the question? Maybe using
Unparseable latex formula:

\text{\cosec}\, 2x = (1 + \cot^2 x)^{1/2}

might work if you know the series for cot⁡\cot.
(edited 7 years ago)
Original post by Zacken
Can you post a picture of the question? Maybe using cosec⁡2x=(1+cot⁡2x)1/2\cosec 2x = (1 + \cot^2 x)^{1/2} might work if you know the series for cot⁡\cot.


I only know the series expansion for cosx and sinx as far as trig is concerned so far.
Q17 part b

ImageUploadedByStudent Room1469987850.092453.jpg


Posted from TSR Mobile
(edited 7 years ago)
Original post by RDKGames
I only know the series expansion for cosx and sinx as far as trig is concerned so far.

ImageUploadedByStudent Room1469987850.092453.jpg


Posted from TSR Mobile


Ah, okay. Then it's as simple as sin⁡2xcsc⁡2x=1\sin 2x \csc 2x = 1, expand sin⁡2x\sin 2x and then compare coefficients.

i.e:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \left(2x - \frac{4x^3}{3} + \frac{4x^5}{15}\right) (a + bx + cx^2 + dx^3) = 1\end{equation*}



dropping terms of degree >3>3.

Also, would make the numbers nicer if you just used sin⁡xcsc⁡x=1\sin x \csc x = 1 and then x↦2xx \mapsto 2x.
Original post by RDKGames
I only know the series expansion for cosx and sinx as far as trig is concerned so far.
Q17 part b

ImageUploadedByStudent Room1469987850.092453.jpg


Posted from TSR Mobile


You need to use part a.

You know the expansion of sin2x so 1/sin2x = (sin2x)^-1 = (.....)^-1. Now use binomial expansion.
(edited 7 years ago)
Original post by Zacken
Ah, okay. Then it's as simple as sin⁡2xcsc⁡2x=1\sin 2x \csc 2x = 1, expand sin⁡2x\sin 2x and then compare coefficients.

i.e:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \left(2x - \frac{4x^3}{3} + \frac{4x^5}{15}\right) (a + bx + cx^2 + dx^3) = 1\end{equation*}



dropping terms of degree >3>3.

Also, would make the numbers nicer if you just used sin⁡xcsc⁡x=1\sin x \csc x = 1 and then x↦2xx \mapsto 2x.


That makes sense, but what coefficients do I exactly compare? If 2ax=1→a=12x2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.
Original post by RDKGames
That makes sense, but what coefficients do I exactly compare? If 2ax=1→a=12x2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.


Look at my post. I remember doing this question 2 year ago.
Original post by Math12345
Look at my post. I remember doing this question 2 year ago.


I attempted binomial expansion but it got too hairy when it came to replacing sin(2x) by its expansion.
Original post by RDKGames
I attempted binomial expansion but it got too hairy when it came to replacing sin(2x) by its expansion.


sin(2x)=2x-(4/3)x^3+... (You can fill in the missing third term)

(sin(2x))^-1 = (2x-(4/3)x^3))^-1 = (2x(1-(2/3)x^2))^-1 = 1/(2x) [1+(2/3)x^2 +...]

You can continue (I used the expansion (1-y)^-1 above).
(edited 7 years ago)
Original post by RDKGames
That makes sense, but what coefficients do I exactly compare? If 2ax=1→a=12x2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.


Math12345's answer is rather nicer, but anyways, for completeness:

Coefficient of x2x^2 gives 2b−12⋅43=0⇒b=132b - \frac{1}{2}\cdot \frac{4}{3}= 0 \Rightarrow b=\frac{1}{3}

x3x^3 gives 2c=0⇒c=02c = 0 \Rightarrow c=0

x4x^4 gives 2d−4b3+215=0⇒d=7452d - \frac{4b}{3} + \frac{2}{15} = 0\Rightarrow d = \frac{7}{45}

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you