July 15, 2020, 09:45:51 pm

### AuthorTopic: First Year University Mathematics Questions  (Read 5210 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

• Posts: 983
• Graphing is where I draw the line.
• Respect: +513
##### First Year University Mathematics Questions
« on: September 10, 2017, 11:04:09 pm »
+10
Here's a place to ask all your first year uni mathematics-related questions!
I couldn't find any others so I thought it would be a good idea to make one
Completed VCE 2016
2015: Biology
2016: Methods | Physics | Chemistry | Specialist Maths | Literature
ATAR : 97.90
2017: BSci (Maths and Engineering) at MelbUni
Feel free to pm me if you have any questions!

• Posts: 983
• Graphing is where I draw the line.
• Respect: +513
##### Re: First Year University Mathematics Questions
« Reply #1 on: September 10, 2017, 11:06:41 pm »
0
I'll start it off with a question of my own:
Do the vectors in the basis of the row space of a matrix + the vectors in the basis of the nullspace / solution space of a matrix make up a basis of Rn, where n is the number of columns in the matrix?
Completed VCE 2016
2015: Biology
2016: Methods | Physics | Chemistry | Specialist Maths | Literature
ATAR : 97.90
2017: BSci (Maths and Engineering) at MelbUni
Feel free to pm me if you have any questions!

#### Sine

• National Moderator
• ATAR Notes Legend
• Posts: 4275
• Respect: +1368
##### Re: First Year University Mathematics Questions
« Reply #2 on: September 10, 2017, 11:12:26 pm »
0
I'm first year uni but doing a 2nd year maths unit LOL, i'll probably be using this thread soon. Great idea Shadowxo

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #3 on: September 10, 2017, 11:36:19 pm »
+11
I'll start it off with a question of my own:
Do the vectors in the basis of the row space of a matrix + the vectors in the basis of the nullspace / solution space of a matrix make up a basis of Rn, where n is the number of columns in the matrix?
$\text{This would impose the condition that the matrix in question, say, }A\\ \text{is a square matrix.}\\ \text{In general, a matrix need not map vectors from }\mathbb{R}^n\text{ to }\mathbb{R}^n.\text{ We may have }\mathbb{R}^m \to \mathbb{R}^n.$
This can be achieved by simply changing the dimensions of the matrix.
$\text{In addition, quite often the vectors of the row space are written down}\\ \text{as row vectors. Hence, if we desire a basis,}\\ \text{we should exploit the transpose.}$
$\text{Therefore, let }A = \begin{pmatrix}\textbf{v}_1^T \\ \vdots \\ \textbf{v}_n^T\end{pmatrix}\\ \text{where }\textbf{v}_k\in \mathbb{R}^n\qquad \forall k=1,\dots,n\text{ are column vectors}$
________________________________________
$\text{The null-space, however, is the set of vectors }\textbf{x}\text{ such that}\\ A\textbf{x} = \begin{pmatrix}\textbf{v}_1\cdot \textbf{x}\\ \vdots \\ \textbf{v}_n \cdot \textbf{x}\end{pmatrix} = \textbf{0}\\ \text{where the dot product formula can be proven by expanding }\textbf{x}\text{ as the }n\text{-tuple vector, and same for }\textbf{v}_k^T$
$\text{And at the same time, the row-space is the set of linear combinations of}\\ S = \{ \textbf{v}_1, \dots, \textbf{v}_n \}.$
$\text{Because }\textbf{v}_k \cdot \textbf{x} = \textbf{0}\\ \text{for some arbitrary vector in the null-space}\\ \text{and any row vector in the matrix}\\ \text{it follows that any vector in the null-space is orthogonal to that in the row-space.}$
$\text{But by extension, because we know that the dot product is a linear operator}\\ \text{this affirms that EVERY linear combination of the vectors in }S\\ \text{i.e. EVERY vector in the row-space is orthogonal to those in the null-space}$
$\text{Hence, it follows that the row space and the null space}\\ \text{are orthogonal complements of each other.}$
________________________________________
$\text{But a very standard linear algebra theorem says that}\\ \text{for any subspace }S, \, S \oplus S^\perp = V\\ \text{where }S^\perp\text{ is the orthogonal complement}\\ \text{and }V\text{ is the original vector space.}$
$\text{From properties of direct sums}\\ \text{this immediately implies that the union of the bases for}\\ \text{the row space and the column space}\\ \text{form a basis for }\mathbb{R}^n.$
Handwavy - Some results are assumed trivial and left as an exercise. Also potentially poorly explained with my 11:36PM dead brain.
« Last Edit: September 11, 2017, 04:25:45 am by RuiAce »

#### AngelWings

• Moderator
• Part of the furniture
• Posts: 1976
• "Angel wings, please guide me..."
• Respect: +1021
##### Re: First Year University Mathematics Questions
« Reply #4 on: December 07, 2017, 09:40:42 pm »
0
Started elsewhere:
Spoiler
Still need help with Taylor series/ approximations here. Learnt it back in first year maths (MTH1030) and need to revise this. Forgotten most of it. Mostly I just need a proof and how it works again. I've also forgotten mostly about limits, so yeah... that'd be great if you could help.
Which parts were you expected to prove? The existence of the $k$-th order Taylor expansion, or that if remainder -> 0 then f is represented by the Taylor series?
To be honest, I’ve forgotten some of the basics. The most common one in these books, after a bit of dissecting, incorporates Taylor series on e^x, giving approximately 1 + x + x^2 + x^3 +... Not so sure how we got from Point A to B and would just like to see how to do it again, how we can prove this and so forth.
$\text{But that's the series for }\frac{1}{1-x}.\\ \text{The Taylor series for }e^x\text{ is }1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
Did you mix some of them up?
That is entirely possible considering the only line of working I have been given is: e^x is approximately 1 + x where any term of order 2+ is ignored because it’d be ridiculously small and thus negligible. (X is meant to be tiny e.g. 10^- 8, hence why it’d be negligible. At least in the context where I got these from - a genetics book. See below.)
Intending on a theoretical genetics project for Honours, which involves some first year math, parts of which my memory stalls on. After previous experience, my intended supervisor advised that during this break, I should go through two genetics books. Both of them indirectly expect you to use Taylor approximations, which I can't remember how it works or how to do them. Hence the revision.

I got that answer because my notes are old and written. I must've been haphazardly copying them down quickly during the lectures. Must've missed the factorials.
Still not quite so sure how we would get the one your wrote for e^x above though, but maybe it's because I've forgotten large chunks of content.
VCE: Psychology | English Language | LOTE | Mathematical Methods (CAS) | Further Mathematics | Chemistry
Uni: (Hons)
University Virtual Events and Open Days

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #5 on: December 07, 2017, 10:00:25 pm »
+4
Started elsewhere:
Spoiler
That is entirely possible considering the only line of working I have been given is: e^x is approximately 1 + x where any term of order 2+ is ignored because it’d be ridiculously small and thus negligible. (X is meant to be tiny e.g. 10^- 8, hence why it’d be negligible. At least in the context where I got these from - a genetics book. See below.)
I got that answer because my notes are old and written. I must've been haphazardly copying them down quickly during the lectures. Must've missed the factorials.
Still not quite so sure how we would get the one your wrote for e^x above though, but maybe it's because I've forgotten large chunks of content.
$\text{Well ok, when it comes to approximations}\\ \text{you just approximate using whatever's reasonable enough.}$
\text{As for the series itself}\\ \text{perhaps the series of the following four functions should be memorised:}\\ \begin{align*}\frac{1}{1-x}&=1+x+x^2+x^3+\dots\\ e^x&=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\ \sin x &= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\\ \cos x &= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots\end{align*}
Optionally you may also memorise the one for $\ln(1-x)$, but that can be obtained by integrating $\frac1{1-x}$.
Remark on first order approximations
This means, that for small $x$, the following are reasonable:
\begin{align*}e^x&\approx 1+x\\ \sin x &\approx x\\ \cos x&\approx 1\end{align*}

These four are generally regarded as the most useful. The first one really isn't anything fancy as it's literally just the geometric series, but exponentials and sinusoids appear quite common naturally. All the other stuff either stem out of these (e.g. $\sinh$), are inverses of these (e.g. $\tan^{-1}$) or are just random shit mathematicians use for convenience.
$\text{With that being said, the actual computation of a Taylor series}\\ \text{is just done by quoting and using the formula.}$
$\text{In general, a Taylor polynomial about a point }x=a\text{ takes the form}\\ P_n(x)= \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k.\\ \text{Assuming the limit exists as }n\to \infty,\text{ a Taylor series is what happens when we do exactly that.}\\ \text{i.e. }P_{\infty}(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$
We omit justification as to why that is the case for the sake of mere computations.
$\text{If the series about }x=0\text{ exists, i.e. taking }a=0,\\ \text{then }f\text{ is }\textit{exactly}\text{ represented by this series.}\\ \text{That is to say, }\boxed{f(x) = \sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k}$
So all we really need to compute is $f(0), f^\prime(0), f^{\prime\prime}(0), f^{\prime\prime\prime}(0), f^{(4)}(0)$ and pretty much, all of the derivatives of $f$, evaluated at 0.
The actual computation starts here.
$\text{But if }f(x) = e^x\text{ then this is easy.}\\ \text{We know that the derivative of }e^x\text{ is still }e^x.\\ \text{Hence, the first, second and so on derivatives must STILL be }e^x,\\ \text{i.e. }\boxed{f^{(k)}(x)=e^x}$
$\text{Which consequently implies }f^{(k)}(0)=e^0=1\\ \text{Hence, }\boxed{e^x = \sum_{k=0}^\infty \frac{1}{k!}x^k}$
Expanding this sum out gives $1 + \frac{x}{1!}+\frac{x^2}{2!}+\dots$ as required.

Small note - The factorials basically appear because they're actually a part of the Taylor series formula.
« Last Edit: December 07, 2017, 10:14:44 pm by RuiAce »

#### legorgo18

• Trendsetter
• Posts: 168
• A future lawyer.
• Respect: +2
##### Re: First Year University Mathematics Questions
« Reply #6 on: March 10, 2018, 02:15:10 pm »
0
Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!
HSC 2017: Advanced English(94), 2U Maths(97), 3U Maths(49), Bio(91), Chem(88), Chinese in context(88)

Atar: 97.55

Studying a bachelor of  actuarial studies/ bachelor of laws at UNSW

Tutoring details: https://highschooltutors.com.au/tutor/12153

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #7 on: March 10, 2018, 02:32:10 pm »
+5
Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!
$\text{One way of doing 32b is to interpret the problem like so:}\\ \text{Take }x\text{ to be some real number. If we want }y \in (x, \infty)\\ \text{then we really just want some }y\text{, to be }\textbf{greater than }x.$
\text{That is, subject to the condition }y^2 - x^2 < 1.\\ \text{Try thinking about what's going on here. We actually have two inequalities now:}\\ \begin{align*}y &> x\\ y &< \sqrt{1+x^2} \end{align*}
Note that I took the positive square root here I'll take the negative square root for 32c. The intuition behind this is that for 32c, I actually want $y \in (-\infty, x)$, i.e. $y < x$, so I'd prefer the negative square root for that one. But here I want $y > x$, and hence i prefer the positive square root here.
$\text{So now I'm just gonna sketch the regions.}$

$\text{Note that if I want both of those inequalities to hold, }\textbf{for all}\, x \in \mathbb{R},\\ \text{I just need to check if for every }x\text{, there's some region that overlaps.}$
And if you look closely, there will always be a bit of the red region overlapping with the green region.
$\text{Thus, this choice of }y\text{ should exist.}$
____________________________________________________________________
$\text{Alternatively, I will also present an algebraic proof.}\\ \text{Given my two inequalities above, I can find an }\textbf{explicit choice}\text{ for }y\text{ to make this work.}$
$\text{I have the inequality }y^2 < 1+x^2.\\ \text{I also have the inequality }y > x\text{, which implies }y^2 > x^2.\\ \text{I can combine these two inequalities to give}\\ \boxed{x^2 < y^2 < 1+x^2}$
$\text{I will simply choose }\textbf{any}\text{ value of }y\text{ that works,}\\ \text{for example, }y^2 = \frac12 + x^2.\\ \text{Which then implies that }\boxed{y = \sqrt{\frac12 + x^2}}$
$\text{And pretty much, that's it. This is colloquially called "proof by construction"}\\ \text{where you }\textbf{give a concrete example}\text{ to prove a }\textbf{"there exists"}\text{ statement.}$

I'll let you try doing part c now whilst I look at the other questions.

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #8 on: March 10, 2018, 03:03:06 pm »
+5
I'll get back to 33. That one is probably the hardest of the bundle you mentioned.
\text{In 35b, we have}\\ \begin{align*}\sup_{n\in \mathbb{N}} \left( \frac12 + \frac14 + \dots + \frac1{2^n} \right)&= \sup_{n\in\mathbb{N}} \frac12 \left( \frac{1-2^{-n}}{1-2^{-1}}\right)\tag{geometric series formula}\\ &= \sup_{n\in \mathbb{N}} (1-2^{-n})\end{align*}
$\text{So the set in which we want to find a least upper bound for is the set}\\ \left\{ x \in \mathbb{R} : x = 1-2^{-n} : n \in \mathbb{N} \right\}\\ \text{which is essentially the supremum of the sequence}\\ \left(\frac12, \frac34, \frac78, \frac{15}{16}, \dots \right)$
Now, by inspection this will just be 1. But let's argue this properly.
$\text{From high school maths, you know that the function }f(x) = 2^{-x}\text{ is monotonic increasing.}\\ \text{Hence, the sequence }b_n = 2^{-n}\text{ is also monotonic decreasing.}$
$\text{But therefore }- 2^{-n}\text{ is monotonic increasing, and thus}\\a_n = 1- 2^{-n}\text{ is also monotonic increasing.}\\ \text{Since this holds }\textbf{for all}\, n \in \mathbb{N},\text{ by this logic,}\\ \text{the supremum of this set should be the limit of }a_n,\\ \text{as }n\to \infty.$
$\therefore \sup_{n \in \mathbb{N}} (1-2^{-n}) = \lim_{n\to \infty} 1-2^{-n} = 1 - 0 = 1.$
_________________________________________________________________________
$\text{On the other hand, this sequence is less easy to consider.}\\ \therefore \text{Let }f : (0,\infty ) \to \mathbb{R}\text{ be the function given by}\\f(x) = \frac{\ln x}{x^{1/3}}$
\text{The quotient rule gives}\\ \begin{align*}f^\prime (x) &= \frac{\frac{x^{1/3}}{x} - \frac13 x^{-2/3} \ln x}{x^{2/3}}\\ &= \frac{3-\ln x}{3x^{4/3}}\end{align*}
$\text{Setting }f^\prime(x) = 0\text{ reveals the stationary point at }\left( e^3, 3e^{-1} \right).\\ \text{Furthermore, this stationary point can be proven to be a local maximum.}\\ \text{Since this is the }\textbf{only}\text{ stationary point, I can plot just the region }\textbf{around }x = e^3.$
(Note that I don't know what happens as $x \to \infty$, nor do I know what happens as $x \to 0^+$. GeoGebra does, but I don't.)

$\text{Now what does that mean?}\\ \text{Note that the above reasoning should suggest that }\left( e^3, 3e^{-1}\right)\\ \text{should be the }\textbf{absolute maximum}\text{ of }y = f(x).$
$\text{This means, that }f(x) \le 3e^{-1}\textbf{ for all }x\in \mathbb{R}.\\ \text{You can't get any higher than }3e^{-1}\text{ along this curve.}\\ \text{So for the function }f(x)\text{, the supremum, which is also the maximum here, will be }3e^{-1}.$
$\text{Now what does that say for }a_n = \frac{\ln n}{n^{1/3}}?$

$\text{Because }a_n = f(n)\text{, just evaluated at the integers and not over all of }\mathbb{R},\\ \text{the max of the sequence should be something similar to the max of the function.}$
$\text{However not quite, since }e^3\text{ is not an integer!}\\ \text{Note that from the calculator, }20 < e^3 < 21.\\ \text{Thus, the max of the }\textbf{sequence}\text{, instead of the function, can only be at }n=20\text{ or }n=21.$
i.e. since the max of the function does not occur at an integer, the max of the sequence must occur at one of the two integers, bordering that original number.
$\text{Plugging into the calculator, we see that}\\ a_{20} \approx 1.1036\text{ and }a_{21}\approx 1.1035\\ \text{Thus, the max of the sequence, and consequently the sup of the sequence, is attained}\\ \text{at }n=20\text{, and hence }\sup_n \frac{\ln n}{n^{1/3}} = \frac{\ln 20}{20^{1/3}}$
(Side note: If the max exists, then the max must equal the sup. Notice how in the last problem the max didn't exist, because we asymptotically approached 1. Here, $f(x)$ didn't just approach $3e^{-1}$, but it actually hit it.)
« Last Edit: March 10, 2018, 03:08:54 pm by RuiAce »

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #9 on: March 10, 2018, 06:53:06 pm »
+6
Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!
$\text{Ok, so for Q33, we first recall the definitions of monotonic increasing and decreasing.}\\ \text{A function }f: [a,b]\to \mathbb{R}\text{ is monotonic increasing if for any }x_1, x_2\in [a,b],\\ \text{If }x_1 < x_2\text{, then }f(x_1) < f(x_2).$
$\text{Similarly, }f:[a,b]\to \mathbb{R}\text{ is monotonic decreasing if for any }x_1,x_2\in [a,b],\\ \text{If }x_1 < x_2, \text{ then }f(x_1) > f(x_2).$
This may come as a surprise. In high school, you were taught that $f$ is a monotonic increasing function, if for any $x \in (a,b)$ it satisfied $f^\prime (x) \ge 0$, where $f^\prime (x)$ is the derivative. And similarly, $\le$ for decreasing. This is actually not a definition, but a theorem. The proof of this theorem requires a neat-ass formula known as the "mean value theorem", which you will learn later on.

But the idea is, we will borrow the theorem to "prove" something is monotonic decreasing, but then relate back to the actual definition.
________________________________________________
$\text{Let }f(t) = \frac{\ln t}{t} \implies f^\prime (t) = \frac{1 - \ln t}{t^2}.\\ \text{Observe that }f^\prime(t) \le 0\text{ whenever }t \le e.\\ \text{Hence }f\text{ is monotonic decreasing for all }t\in [e, \infty).$
$\text{Now, let }x\in [e, \infty).\\ \text{Take }x_1 = x\text{ and }x_2 = x+1.\text{ From the definition of monotonic decreasing,}\\ \text{since we know that }x_1 < x_2\text{, we also know that }f(x_1) >f(x_2).$
\text{Hence,}\\ \begin{align*}\frac{\ln x}{x} &> \frac{\ln (x+1) }{x+1}\\ \implies (x+1) \ln x &> x \ln (x+1) \end{align*}
which, coincidentally, means that $x$ is NOT in our set. Because our set is the set of all $p$ such that $(p+1) \ln x < p \ln (p+1)$. Note that I'm just using $p$ to not mix up three different $x$'s in the same question.
$\text{Therefore, any real number }x\text{, that is greater than or equal to }e,\\ \text{is }\textbf{guaranteed}\text{ to be not in our set.}\\ \text{This means, only elements }\textbf{less}\text{ than }e\text{ can be in our set.}$
$\text{Thus, }e\text{ is an example of an upper bound for our set.}$
Remark: $e$ may not necessarily be the least upper bound, i.e. the supremum. In fact, the supremum is actually approximately at 2.293, which is certainly smaller than 2.718

Also - the lower bound for this set is pretty much 0, because you can't have negative elements in that set to begin with. (Reason - the domain restriction of the log function)
« Last Edit: March 10, 2018, 08:56:04 pm by RuiAce »

#### Mackenzie2000

• Fresh Poster
• Posts: 1
• Respect: 0
##### Re: First Year University Mathematics Questions
« Reply #10 on: April 05, 2018, 08:07:21 am »
+1
When working with matrices, and determining their geometric description, how do we know whether they are parallel, the same, or have the planes intersect in either a line or at a single point?
I am referring to both questions with three equations in three variable, and questions with two equations, but three variables.

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #11 on: April 05, 2018, 08:45:43 am »
+5
When working with matrices, and determining their geometric description, how do we know whether they are parallel, the same, or have the planes intersect in either a line or at a single point?
I am referring to both questions with three equations in three variable, and questions with two equations, but three variables.

Essentially, the geometric interpretation of the solutions (and hence the original planes) depends on the nature of the solutions we had found.

In the 3 equation, 3 variable case, we are considering 3 planes in $\mathbb{R}^3$. So we have the following:
- There is no solution. When this is the case, we're saying that the 3 planes have no common point of intersection. This could be because they appear to form a triangular prism between them, or because at least 2 out of the 3 planes are parallel to one another.
- There is one unique solution. When this is the case, the planes are essentially intersecting at a common point. An easy example of this is just the x-y, y-z and x-z planes, which only intersect at (0,0,0).
- There are infinite solutions. Infinite solutions are characterised by the number of parameters there are in our solution.
- If there is only one parameter, say $t$, then the solutions are of the form $\textbf{x} = a + t\textbf{v}$. This represents a line in $\mathbb{R}^3$, passing through the point $A$ and in the direction of the vector $\textbf{v}$. In general, the intersection forms a lie when the planes are just rotations of one another, but all 3 of them aren't the same plane as in the case below.
- If there are two parameters, say $\lambda$ and $\mu$, then the solutions are of the form $\textbf{x} = \textbf{a}+ \lambda \textbf{u} + \mu \textbf{v}$. This now represents a plane in $\mathbb{R}^3$, which passes through $A$ and is spanned by $\textbf{u}$ and $\textbf{v}$. However, this is only possible if the three planes at the start actually all coincided with one another.

I find that figure 1.1.2 of this article gives a good visual representation of them.

On the other hand, when we have 2 equations, we're simply saying we now have only 2 planes in $\mathbb{R}^3$. Because there's only 2 planes, our possibilities are severely narrowed
- The only way there can be no solutions is if the planes are parallel to one another.
- There can never be a unique solution. (Why?) If two planes in $\mathbb{R}^3$ are not parallel then they must have infinite solutions.
- If there are two parameters, then once again the planes coincided.
- If there is only one parameter, again the intersection is a line. This is essentially every case when the planes aren't parallel with one another.

#### sophieren

• Fresh Poster
• Posts: 1
• Respect: 0
##### Re: First Year University Mathematics Questions
« Reply #12 on: April 05, 2018, 11:19:33 am »
0
Hi,
I was wondering if you are able to solve these questions for me? It would be greatly appreciated!

1) In the triangle ABC, AB = sqrt(2), AC = 1/sqrt(2) and the angle at A is 60◦. Find the length of BC and the size of the angle at C.
2) If sin A = 3/5 where 90◦ < A < 180◦ and cos B = 5/13 where 0◦ < B < 90◦, find sin(A − B)
3) In the triangle ABC, AB = 2, BC = 1 + sqrt(3) and the angle at B is 30◦, Find the length of AC and the size of the angle at C.
4) If sin A = 5/13 where 90◦ < A < 180◦ and cos B = 3/5 where 0◦ < B < 90◦, find sin(A − B).

Thanks!

#### TrueTears

• TT
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 16369
• Respect: +658
##### Re: First Year University Mathematics Questions
« Reply #13 on: April 05, 2018, 02:38:50 pm »
+3
1) is just a straightforward application of the cosine rule. 2) and 4) are applications of the sine difference rule. Have you had a go at applying those?
Currently studying: PhD in economics at MIT.

Interested in financial economics, econometrics, and asset pricing.

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8745
• "All models are wrong, but some are useful."
• Respect: +2523
##### Re: First Year University Mathematics Questions
« Reply #14 on: April 05, 2018, 02:47:37 pm »
+3
Hi,
I was wondering if you are able to solve these questions for me? It would be greatly appreciated!

1) In the triangle ABC, AB = sqrt(2), AC = 1/sqrt(2) and the angle at A is 60◦. Find the length of BC and the size of the angle at C.
2) If sin A = 3/5 where 90◦ < A < 180◦ and cos B = 5/13 where 0◦ < B < 90◦, find sin(A − B)
3) In the triangle ABC, AB = 2, BC = 1 + sqrt(3) and the angle at B is 30◦, Find the length of AC and the size of the angle at C.
4) If sin A = 5/13 where 90◦ < A < 180◦ and cos B = 3/5 where 0◦ < B < 90◦, find sin(A − B).

Thanks!

Essentially Q1 and Q3 are just high school trigonometry - the main thing you require is the cosine rule (and possibly the sine rule.) Since Q4 follows the exact same as Q2, I will only do Q2.
$\text{Note that }A\text{ is a second quadrant angle.}\\ \text{For this reason, }\sin A > 0\text{ and }\cos A < 0.$
$\text{Draw a triangle with hypotenuse 5 and opp side 3.}\\ \text{Then the adj side has length 4,}\\ \text{so }\cos A = -\frac{4}{5}.$
$\text{On the other hand }B\text{ is a first quadrant angle, so both }\sin B > 0\text{ and }\cos B > 0.\\ \text{Drawing another triangle, we see that }\sin B = \frac{12}{13}.$
\text{From here, simply expand the compound angle and substitute in.}\\ \begin{align*}\sin (A-B) &= \sin A \cos B - \cos A \sin B\\ &= \frac35 \times \frac5{13} +\frac45\times\frac{12}{13}\\ &= \frac{63}{65}\end{align*}