Author Topic: Specialist 1/2 Question Thread! (Read 121128 times) Tweet Share

Syndicate · « **Reply #45 on:** July 07, 2016, 10:52:02 am »

Quote from: anotherworld2b on July 06, 2016, 10:56:14 pm

Thank you so much for your help
I wanted to ask some questions if that's okay
For Q10
How did It go from
$=\sqrt2 (sinx-cosx)(sinxcos45+sin45cosx)$
to
$= \sqrt2 (sinx-cosx)( \frac{sinx+cosx}{ \sqrt2 } )$

For these kind off questions, I would recommend you to remember the exact values of trigonmetric functions and its identities. So basically:
$sin45 = cos45 = \frac{1}{\sqrt2}$

I basically coverted cos 45 and sin 45 into its exact value (which was 1/sqrt(2)), and as they had the same denominator, I was able to put sinx/sqrt(2) and cosx/sqrt(2) as one fraction.

I would really recommend you to remember these^

Quote from: anotherworld2b on July 06, 2016, 10:56:14 pm

$= sin^2x-cos^2x$
$= -1(cos^2x-sin^2x$
$= -cos(2x)$
$= -1(2cos^2x-1)$

Subsequent to multiplying the two equations together, you should end up sin^2x-cos^2x. Then, take out the negative factor, in order for you to get -cos(2x).

cos(2x) has three different variations.
= cos^2x - sin^2x
= 2cos^2x-1
= 1- 2sin^2x

Quote from: anotherworld2b on July 06, 2016, 10:56:14 pm

Further explanation would be greatly appreciated
I am still greatly lacking in understanding how to apply the trig identities and manipulating them
I also wanted to ask about Q11. How did you prove it from this step onwards?
I am still relatively new to trig identities proofs

$\frac{ \frac{sinx+cosx}{cosx} }{\frac{cosx-sinx}{cosx}}$
$\frac{sinx+cosx}{cosx} \times \frac{cosx}{cosx-sinx}$
$\frac{sinx+cosx}{cosx-sinx}$
$\frac{sinx+cosx}{cosx-sinx} \times \frac{sinx+cosx}{cosx+sinx}$
$\frac{sin^2x+cos^2x+2cosxsinx}{cos^2x-sin^2x}<br />$
$\frac{1+2sinxcosx}{cos2x}$

$\frac{1+2sinxcosx}{1-2sin^2x} = RHS$

Well, when you have fractions situated within fractions, you would multiply the numerator with the reciprocal of the denominator. Hence, when you are multiplying them together, you would begin by canceling out the cosx, in both fractions.

To simplify the fraction, you would multiply the numerator and the denomirator, with the invert of the denominator (which in this case was cosx + sinx).

Then using your knowledge of trig identities, you would instantly convert the sin^2x + cos^2x to 1 (as sin^2x + cos^2x = 1).

Likewise, you would also convert the cos^2x - sin^2x to cos2x, and then to 1 - 2sin^2x, in order to prove what your are given.

anotherworld2b · « **Reply #46 on:** July 10, 2016, 12:52:44 am »

I am still a bit confused
$= sin^2x-cos^2x$
-1 was taken out
$= -1(cos^2x-sin^2x$
what happened here? why is it 2x?
$= -cos(2x)$
$= -1(2cos^2x-1)$

RuiAce · « **Reply #47 on:** July 10, 2016, 10:31:09 am »

Quote from: anotherworld2b on July 10, 2016, 12:52:44 am

I am still a bit confused
$= sin^2x-cos^2x$
-1 was taken out
$= -1(cos^2x-sin^2x$
what happened here? why is it 2x?
$= -cos(2x)$
$= -1(2cos^2x-1)$

$\cos 2x = \cos^2x - \sin^2 x\\ \text{is your double angle formula for cosine.}$
$\text{Proof: In the compound angle formula }\cos (A+B)=\cos A \cos B - \sin A \sin B\\ \text{Let }A=B=x$
$\text{Additionally}\\ \begin{align*}-(2\cos^2x-1)&=-(\cos^2x+\cos^2x-1)\\ &=-(\cos^2x + (1-\sin^2x) - 1)\\ &= -(\cos^2x-\sin^2x)\end{align*} \\ \text{so the steps are clearly reversible.}$

Syndicate · « **Reply #48 on:** July 10, 2016, 10:37:25 am »

These are all the double angle formulas you are required to remember.

Clockwork · « **Reply #49 on:** July 20, 2016, 05:50:21 pm »

I have a question on changing Cartesian equations to their polar form and I'm a bit stuck on how to do this one.
The Cartesian equation is $y = \sin{x}$ , so far I've converted it to $r\sin{\theta} = sin({r\cos{\theta}})$ .

I have a feeling I'm way off with this question, could someone please provide some advice with it? If I have somehow done it correctly so far, can I simpilify the equation any further than this?

RuiAce · « **Reply #50 on:** July 20, 2016, 05:55:31 pm »

Quote from: Clockwork on July 20, 2016, 05:50:21 pm

I have a question on changing Cartesian equations to their polar form and I'm a bit stuck on how to do this one.
The Cartesian equation is $y = \sin{x}$ , so far I've converted it to $r\sin{\theta} = sin({r\cos{\theta}})$ .

I have a feeling I'm way off with this question, could someone please provide some advice with it? If I have somehow done it correctly so far, can I simpilify the equation any further than this?

Are you sure that this was the question

Because that's about as simplified as it can get.

Clockwork · « **Reply #51 on:** July 20, 2016, 06:02:46 pm »

Quote from: RuiAce on July 20, 2016, 05:55:31 pm

Are you sure that this was the question

Because that's about as simplified as it can get.

Thanks for that! Yeah, that was the question I was given. I just wasn't 100% sure on the answer because it seemed a bit 'messy' compared to the other questions we've done so far.

RuiAce · « **Reply #52 on:** July 20, 2016, 06:06:54 pm »

Quote from: Clockwork on July 20, 2016, 06:02:46 pm

Thanks for that! Yeah, that was the question I was given. I just wasn't 100% sure on the answer because it seemed a bit 'messy' compared to the other questions we've done so far.

$\text{We can analyse the reason why we cannot express }r\text{ as a function of }\theta\text{ graphically.} \\ \text{Suppose we chose }\theta = \arctan \frac{1}{10}\\ \text{Sketch }y=\frac{x}{10}\text{ and }y=\sin x\text{ to an accurate scale (or use software).}$
$\text{From the graph, clearly the line }y=\frac{x}{10}\text{ intersects the sine curve twice (in the first quadrant). This means that}\\ \text{when }\theta=\arctan \frac{1}{10}\text{, there exists two values for }r\\ \text{This defies the definition of a function: One input has only one output.}$
$\text{Hence the best we can get is relating }r\text{ as an implicit function of }\theta$

Adequace · « **Reply #53 on:** July 20, 2016, 09:30:31 pm »

Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks

RuiAce · « **Reply #54 on:** July 20, 2016, 09:39:22 pm »

Quote from: Adequace on July 20, 2016, 09:30:31 pm

Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks

$\text{I'm not sure how the VCE teaches it but the shortest distance is }\\ \left|\text{proj}_{\overrightarrow{AC}}\overrightarrow{AB} - \overrightarrow{AB} \right|\\ \because A\text{ lies on the line }AC\quad \text{(note: }BC\text{ should've also worked however.)}$
$\\ \text{where }\text{proj}_{\overrightarrow{AC}}\overrightarrow{AB}=\frac{\overrightarrow{AC}\cdot \overrightarrow{AB}}{\overrightarrow{AC}\cdot \overrightarrow{AC}}\overrightarrow{AC}\\ \text{which is seemingly what you found in b)}$

Edit: Yes, this is a formula. If you want me to prove it that will be a slight bit harder.

Syndicate · « **Reply #55 on:** July 20, 2016, 10:19:41 pm »

Quote from: Adequace on July 20, 2016, 09:30:31 pm

Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks

The shortest distance from point B to line AC is when the line between point B to line AC is perpendicular.

In this case, it's the red line^

In order to do your question, you would first find the vector resolute of AB in the direction of AC, (because that gives you the vector AD(assuming D is the point where the line red line meets on the vector AC).

Now in order for you to get the Vector BD (which is the shortest distance from point B to line AC), you would subtract AB by AD. Then find the modulus of BD in order to get the distance of that vector.

Adequace · « **Reply #56 on:** July 21, 2016, 07:36:58 pm »

Thanks Rui and Syndicate

Adequace · « **Reply #57 on:** July 24, 2016, 06:56:47 pm »

Could I have some help with these 2 questions for vector proofs http://m.imgur.com/a/KPQDE

For Q9) How does the linked solution prove that they bisect with their working? They proved that BM = MD, are we supposed to then assume that the other diagonal is just going to bisect it with logic..?

For Q10) I'm completely lost with their working. How did they form those 2 simultaneous equations and what are they actually representing?

Thanks

Syndicate · « **Reply #58 on:** July 24, 2016, 07:26:29 pm »

Quote from: Adequace on July 24, 2016, 06:56:47 pm

Could I have some help with these 2 questions for vector proofs http://m.imgur.com/a/KPQDE

For Q9) How does the linked solution prove that they bisect with their working? They proved that BM = MD, are we supposed to then assume that the other diagonal is just going to bisect it with logic..?

For Q10) I'm completely lost with their working. How did they form those 2 simultaneous equations and what are they actually representing?
Thanks

Q9) Since, AC and BD have the same midpoint, the diagonals of the parallelogram bisect (divide into two equal pieces) each other

Q10) An altitude is a line segment, which is perpendicular to the base.

Since, they all are perpendicular and meet at the point O (assume it's the point where all the lines intersect in the image), you can assume that the dot product of A to BC and B to AC are equal. So here, you basically have to prove that OC (c) is perpendicular to AB (-a+b), by equating the two dot products, and showing that they equal c . (b-a)

RuiAce · « **Reply #59 on:** July 24, 2016, 07:31:03 pm »

Just to add on.

Note that M was defined as the midpoint of AC. Because it is a definition, BM, MD and BD automatically bisects AC.

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Author Topic: Specialist 1/2 Question Thread! (Read 121128 times) Tweet Share

Syndicate

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