Author Topic: VCE Specialist 3/4 Question Thread! (Read 2825770 times) Tweet Share

jazzycab · « **Reply #9195 on:** February 15, 2018, 06:38:08 pm »

Quote from: noregret on February 15, 2018, 05:30:32 pm

Hello, how do I find the sum of all multiples of 6 between 100 and 600
Sum=102+108+114+.....+594
If Un=a+(n-1)d I know the Un=594 and a=102, but I cannot get n and d.

Each multiple of six differs from the previous by exactly six, therefore the common difference, d=6. Sub this in and solve for n

noregret · « **Reply #9196 on:** February 15, 2018, 07:02:08 pm »

Quote from: jazzycab on February 15, 2018, 06:38:08 pm

Each multiple of six differs from the previous by exactly six, therefore the common difference, d=6. Sub this in and solve for n

Thanks for your help and how come Given U1=4 and Un=Un-1+5 find 18 sigma notation r=7 (Ur+3)
{Un}=4,9,14 and how do I solve the question?

jazzycab · « **Reply #9197 on:** February 15, 2018, 08:46:44 pm »

It's really difficult to understand the question you're asking without the appropriate notation. Is it supposed to be:
$U_1=4,U_n=U_{n-1}+5\\\text{Find }\sum\limits_{r=7}^{18}{U_{r+3}}$

If so, we have an arithmetic sequence with initial term 4 and common difference 5, and we're looking for the arithmetic series of the 10th to 21st terms (inclusive).
That is:
$\sum\limits_{r=7}^{18}{U_{r+3}}=U_{7+3}+U_{8+3}+U_{9+3}+...+U_{17+3}+U_{18+3}\\U_{10}+U_{11}+...+U_{20}+U_{21}=\frac{12}{2}\left(4+5\times9+4+5\times20\right)\\=6\times153\\=918$

JamesMaths · « **Reply #9198 on:** February 17, 2018, 12:22:05 am »

Quote from: microwaverdetective on February 14, 2018, 08:11:12 pm

If A and B are acute angles such that sin(A) = 3/5 and cos(B) = 5/13, find tan(A+B) without evaluating A or B.
Working please! Cheers

Could I present my solution here?

Thanks
James.

JamesMaths · « **Reply #9199 on:** February 17, 2018, 12:34:53 am »

Quote from: noregret on February 15, 2018, 05:30:32 pm

Hello, how do I find the sum of all multiples of 6 between 100 and 600
Sum=102+108+114+.....+594
If Un=a+(n-1)d I know the Un=594 and a=102, but I cannot get n and d.

Here is my solution.

Thanks
James

Mattjbr2 · « **Reply #9200 on:** February 25, 2018, 07:09:21 pm »

Hey guys, isn't sinx+cosx=1 <=> (sinx+cosx)^2=1^2? Since A=B <=> A^2=B^2. Why then, do they give differing solutions for x?

I know that you can't just square both sides of root(4)=-2 because you will change the equality. But, isn't sinx+cosx positive and so there isn't any disappearance of any negatives? Or is this ignoring negative values of sinx and cosx, since both are elements of [-1,1]. So, does squaring the LHS of sinx+cosx=1 eliminate the negative portion of the range and thus give us different solutions?

Sine · « **Reply #9201 on:** February 25, 2018, 07:13:45 pm »

Quote from: Mattjbr2 on February 25, 2018, 07:09:21 pm

Hey guys, isn't sinx+cosx=1 <=> (sinx+cosx)^2=1^2? Since A=B <=> A^2=B^2. Why then, do they give differing solutions for x?

(sinx+cosx)^2=1^2 nor is sin(x) + cos(x) = 1 a correct equation at least not for all x values

The pythagorean identity is sin²(x) + cos²(x) = 1

Mattjbr2 · « **Reply #9202 on:** February 25, 2018, 07:16:27 pm »

Quote from: Sine on February 25, 2018, 07:13:45 pm

(sinx+cosx)^2=1^2 nor is sin(x) + cos(x) = 1 a correct equation at least not for all x values

The pythagorean identity is sin²(x) + cos²(x) = 1

Ooops. I think I should've been more clear.
Question: solve for x: sinx+cosx=1, | x is an element of [0,2pi]

I attached my working. Where is my error?

RuiAce · « **Reply #9203 on:** February 25, 2018, 07:21:14 pm »

Quote from: Mattjbr2 on February 25, 2018, 07:16:27 pm

Ooops. I think I should've been more clear.
Question: solve for x: sinx+cosx=1, | x is an element of [0,2pi]

I attached my working. Where is my error?

$\text{Did you make a domain adjustment?}\\ x\in [0,2\pi]\text{ would consequently imply }\boxed{2x \in [0,4\pi]}$
$\text{So }\sin 2x = 0\text{ gives the solutions}\\ 2x = 0, \pi, 2\pi, 3\pi, 4\pi$
(And of course, you have to sub some back into the original equations to do a check. Because you introduced extra solutions when you did the squaring business.

Mattjbr2 · « **Reply #9204 on:** February 25, 2018, 07:23:20 pm »

Quote from: RuiAce on February 25, 2018, 07:21:14 pm

$\text{Did you make a domain adjustment?}\\ x\in [0,2\pi]\text{ would consequently imply }\boxed{2x \in [0,4\pi}$
$\text{So }\sin 2x = 0\text{ gives the solutions}\\ 2x = 0, \pi, 2\pi, 3\pi, 4\pi$
(And of course, you have to sub some back into the original equations to do a check. Because you introduced extra solutions when you did the squaring business.

Check this out. What is this trickery?

How is it giving 2 different sets of solutions for the same equation

RuiAce · « **Reply #9205 on:** February 25, 2018, 07:24:37 pm »

Quote from: Mattjbr2 on February 25, 2018, 07:23:20 pm

Check this out. What is this trickery? How is it giving 2 different sets of solutions for the same equation

Yeah. That's what happens when you square both sides of an equation. You introduce extra solutions that you must sub back in to explicitly discard.
$\text{Essentially the reason why this happens is because}\\ \text{of a little extra basic algebra.}\\ \text{The solutions of }\boxed{f(x) = g(x)}\text{ will be }\textit{some if not all}\text{ of the solutions}\\ \text{of }\boxed{[f(x)]^2=[g(x)]^2}$
$\text{But it }\textbf{won't always be all of them.}\\ \text{This is because the solutions to }[f(x)]^2 = [g(x)]^2\text{ are}\\ f(x)=g(x)\textbf{ AND }\boxed{f(x) = -g(x)}$
$\text{So when you solve }(\sin x + \cos x)^2 = 1\\ \text{you find the solutions to }\sin x + \cos x = 1\\ \textbf{but also the solutions to }\boxed{\sin x + \cos x = -1}$
In your case, $0, \frac\pi2, 2\pi$ are the solutions to your original equation. But when you solved $ (\sin x + \cos x)^2 = 1$, $\pi $ and $ \frac{3\pi}2 $ appeared as well because they're solutions to $\boxed{\sin x + \cos = -1} $. This is why you need to explicitly discard them.

Mattjbr2 · « **Reply #9206 on:** February 25, 2018, 07:30:11 pm »

Quote from: RuiAce on February 25, 2018, 07:24:37 pm

Yeah. That's what happens when you square both sides of an equation. You introduce extra solutions that you must sub back in to explicitly discard.
$\text{Essentially the reason why this happens is because}\\ \text{of a little extra basic algebra.}\\ \text{The solutions of }\boxed{f(x) = g(x)}\text{ will be }\textit{some if not all}\text{ of the solutions}\\ \text{of }\boxed{[f(x)]^2=[g(x)]^2}$
$\text{But it }\textbf{won't always be all of them.}\\ \text{This is because the solutions to }[f(x)]^2 = [g(x)]^2\text{ are}\\ \boxed{f(x)=g(x)}\textbf{ AND }\boxed{f(x) = -g(x)}$
$\text{So when you solve }(\sin x + \cos x)^2 = 1\\ \text{you find the solutions to }\sin x + \cos x = 1\\ \textbf{but also the solutions to }\boxed{\sin x + \cos x = -1}$

Why doesn't the textbook mention this, let alone even hint at it anywhere... TIL. Thank you!

Edit: to your edit: thank you so much! x10 clearer

lzxnl · « **Reply #9207 on:** February 26, 2018, 03:44:19 pm »

For what it's worth, this is my preferred way of solving this problem.

$ \sin(x) + \cos(x) = 1\\ r\sin(x + \alpha) = 1\\ r\sin(x + \alpha) = r\sin(x)\cos(\alpha) + r\cos(x)\sin(\alpha) = \sin(x) + \cos(x)\\ \therefore r\cos(\alpha) = 1, r\sin(\alpha) = 1 $
where I compared coefficients of sine and cosine. Using Pythagorean identity,
$ r^2(\cos^2(\alpha) + \sin^2(\alpha)) = r^2 = 2\\ r = \sqrt{2}\\ \sqrt{2}\cos(\alpha) = \sqrt{2}\sin(\alpha) = 1\\ \alpha = \frac{\pi}{4}\\ \therefore \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1\\ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\\ x + \frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}\\ x = 0, \frac{\pi}{2}, 2\pi $
Bit overkill, but it solves the problem directly with no ambiguity. The whole point of the compound angle formula here was to combine the different sine and cosine terms into one trig expression.

RuiAce · « **Reply #9208 on:** February 26, 2018, 04:19:48 pm »

Quote from: lzxnl on February 26, 2018, 03:44:19 pm

For what it's worth, this is my preferred way of solving this problem.

$ \sin(x) + \cos(x) = 1\\ r\sin(x + \alpha) = 1\\ r\sin(x + \alpha) = r\sin(x)\cos(\alpha) + r\cos(x)\sin(\alpha) = \sin(x) + \cos(x)\\ \therefore r\cos(\alpha) = 1, r\sin(\alpha) = 1 $
where I compared coefficients of sine and cosine. Using Pythagorean identity,
$ r^2(\cos^2(\alpha) + \sin^2(\alpha)) = r^2 = 2\\ r = \sqrt{2}\\ \sqrt{2}\cos(\alpha) = \sqrt{2}\sin(\alpha) = 1\\ \alpha = \frac{\pi}{4}\\ \therefore \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1\\ \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\\ x + \frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}\\ x = 0, \frac{\pi}{2}, 2\pi $
Bit overkill, but it solves the problem directly with no ambiguity. The whole point of the compound angle formula here was to combine the different sine and cosine terms into one trig expression.

It's actually the standard method in HSC MX1. I still don't understand why they don't teach it in the VCE tbh.

noregret · « **Reply #9209 on:** February 26, 2018, 08:53:19 pm »

Hello, how do you solve

Find the sum to n terms (1 times 2)+(3 times 4)+(5 times 6)+... (factorise answer)?

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2825770 times) Tweet Share

jazzycab

Re: Specialist 3/4 Question Thread!

noregret

Re: Specialist 3/4 Question Thread!

jazzycab

Re: Specialist 3/4 Question Thread!

JamesMaths

Re: Specialist 3/4 Question Thread!

JamesMaths

Re: Specialist 3/4 Question Thread!

Mattjbr2

Re: VCE Specialist 3/4 Question Thread!

Sine

Re: VCE Specialist 3/4 Question Thread!

Mattjbr2

Re: VCE Specialist 3/4 Question Thread!

RuiAce

Re: VCE Specialist 3/4 Question Thread!

Mattjbr2

Re: VCE Specialist 3/4 Question Thread!

RuiAce

Re: VCE Specialist 3/4 Question Thread!

Mattjbr2

Re: VCE Specialist 3/4 Question Thread!

lzxnl

Re: VCE Specialist 3/4 Question Thread!

RuiAce

Re: VCE Specialist 3/4 Question Thread!

noregret

Re: VCE Specialist 3/4 Question Thread!

Recent Posts

User:
Password: