Author Topic: Specialist 1/2 Question Thread! (Read 121163 times) Tweet Share

Adequace · « **Reply #60 on:** July 24, 2016, 08:16:01 pm »

Thanks guys again!

Adequace · « **Reply #61 on:** July 24, 2016, 08:41:55 pm »

Quote from: Syndicate on July 24, 2016, 07:26:29 pm

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

(Image removed from quote.)
Q10) An altitude is a line segment, which is perpendicular to the base.
(Image removed from quote.)
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)

Sorry for double post, but I'm still a bit unsure on why you can assume the dot product of A to BC and B to AC are equal even though they are perpendicular?

Wouldn't vector a and b be different lengths?

Thanks

Syndicate · « **Reply #62 on:** July 24, 2016, 10:03:01 pm »

Quote from: Adequace on July 24, 2016, 08:41:55 pm

Sorry for double post, but I'm still a bit unsure on why you can assume the dot product of A to BC and B to AC are equal even though they are perpendicular?

Wouldn't vector a and b be different lengths?

Thanks

They can equate since they equal the same dot product when you expand them out (the book hasn't shown that)

Personally, I would have done something like:
$a . (c-b) = 0$
$b . (c-a) = 0$
$a.c - a.b = 0$
$b.c-b.a = 0$
$a.c = a.b \space (1)$
$b.c=b.a \space (2)$
$\text{a.b and b.a will yield the same result, hence, you can equate the two equations in order to solve simultaneously}$
$b.c = a.c$
$b.c-a.c =0$
$c.(b-a)=0$
QED

anotherworld2b · « **Reply #63 on:** July 24, 2016, 11:35:01 pm »

Could i get help with these questions?
I've been trying to solve them the past 2 hours TTT ^ TTT

RuiAce · « **Reply #64 on:** July 24, 2016, 11:46:51 pm »

Quote from: anotherworld2b on July 24, 2016, 11:35:01 pm

Could i get help with these questions?
I've been trying to solve them the past 2 hours TTT ^ TTT

Are these proofs?
$\begin{align*}LHS&=\sin 4\theta\\ &= \sin (2\times 2\theta)\\ &= 2\sin2\theta \cos 2\theta\\ &= 2(2\sin\theta\cos\theta)(\cos^2\theta - \sin^2\theta)\\ &= 4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta\\&=RHS\end{align*}$
$\begin{align*}LHS&=\frac{\sin 2\theta - \sin \theta}{1-\cos \theta + \cos 2\theta}\\ &= \frac{2\sin\theta\cos\theta - \sin\theta}{1-\cos\theta + (2\cos^2\theta-1)}\\ &= \frac{\sin \theta(2\cos \theta - 1)}{\cos \theta (2\cos\theta-1)}\\ &=\tan\theta\\ &=RHS\end{align*}$
$\text{What was the motivation to choose }2\cos^2\theta-1\text{ for the second question?} \\ \text{The }-1\text{ that pops out cancels out with the }+1$

anotherworld2b · « **Reply #65 on:** July 25, 2016, 01:36:00 am »

Quote from: RuiAce on July 24, 2016, 11:46:51 pm

Are these proofs?
$\begin{align*}LHS&=\sin 4\theta\\ &= \sin (2\times 2\theta)\\ &= 2\sin2\theta \cos 2\theta\\ &= 2(2\sin\theta\cos\theta)(\cos^2\theta - \sin^2\theta)\\ &= 4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta\\&=RHS\end{align*}$
$\begin{align*}LHS&=\frac{\sin 2\theta - \sin \theta}{1-\cos \theta + \cos 2\theta}\\ &= \frac{2\sin\theta\cos\theta - \sin\theta}{1-\cos\theta + (2\cos^2\theta-1)}\\ &= \frac{\sin \theta(2\cos \theta - 1)}{\cos \theta (2\cos\theta-1)}\\ &=\tan\theta\\ &=RHS\end{align*}$
$\text{What was the motivation to choose }2\cos^2\theta-1\text{ for the second question?} \\ \text{The }-1\text{ that pops out cancels out with the }+1$

Thank you so much for your help i was stuck on them for hours

and yes they are proofs.

anotherworld2b · « **Reply #66 on:** July 31, 2016, 07:22:48 pm »

I was really confused on how to solve this question

zsteve · « **Reply #67 on:** July 31, 2016, 07:31:14 pm »

Quote from: anotherworld2b on July 31, 2016, 07:22:48 pm

I was really confused on how to solve this question

This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)

$\begin{gather*} 2\sin(x) + 5\cos(x) = 3\\ \sqrt{2^2 + 5^2}[\dfrac{2}{\sqrt{2^2+5^2}}\sin(x) + \dfrac{5}{\sqrt{2^2+5^2}}\cos(x)] = 3\\ \sqrt{29}[\dfrac{2}{\sqrt{29}}\sin(x) + \dfrac{5}{\sqrt{29}}\cos(x)] = 3\\ \text{Let } \sin(u) = \dfrac{2}{\sqrt{29}}, u \in [0, \pi/2]\\ \Rightarrow \cos(u) = \sqrt{1-\sin^2(u)} = \dfrac{5}{\sqrt{29}} \text{ verify this yourself}\\ \Rightarrow \sqrt{29}[\sin(u)\sin(x) + \cos(u)\cos(x)] = 3\\ \Rightarrow \sin(u)\sin(x) + \cos(u)\cos(x) = \dfrac{3}{\sqrt{29}}\\ \Rightarrow \cos(x-u) = \dfrac{3}{\sqrt{29}} \text{ using compound angle formula}\\ \text{But } \sin(u) = \dfrac{2}{\sqrt{29}} \Rightarrow u = \arcsin \dfrac{2}{\sqrt{29}} \approx 0.3805\\ \cos(x-0.3805) = \dfrac{3}{\sqrt{29}}, 0\le x\le 2\pi \end{gather*}$

Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.

RuiAce · « **Reply #68 on:** July 31, 2016, 07:51:41 pm »

Do they teach the tangent half-angle substitution or the auxiliary angle method in this course?

Gogo14 · « **Reply #69 on:** July 31, 2016, 07:55:25 pm »

Quote from: zsteve on July 31, 2016, 07:31:14 pm

This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)

$\begin{gather*} 2\sin(x) + 5\cos(x) = 3\\ \sqrt{2^2 + 5^2}[\dfrac{2}{\sqrt{2^2+5^2}}\sin(x) + \dfrac{5}{\sqrt{2^2+5^2}}\cos(x)] = 3\\ \sqrt{29}[\dfrac{2}{\sqrt{29}}\sin(x) + \dfrac{5}{\sqrt{29}}\cos(x)] = 3\\ \text{Let } \sin(u) = \dfrac{2}{\sqrt{29}}, u \in [0, \pi/2]\\ \Rightarrow \cos(u) = \sqrt{1-\sin^2(u)} = \dfrac{5}{\sqrt{29}} \text{ verify this yourself}\\ \Rightarrow \sqrt{29}[\sin(u)\sin(x) + \cos(u)\cos(x)] = 3\\ \Rightarrow \sin(u)\sin(x) + \cos(u)\cos(x) = \dfrac{3}{\sqrt{29}}\\ \Rightarrow \cos(x-u) = \dfrac{3}{\sqrt{29}} \text{ using compound angle formula}\\ \text{But } \sin(u) = \dfrac{2}{\sqrt{29}} \Rightarrow u = \arcsin \dfrac{2}{\sqrt{29}} \approx 0.3805\\ \cos(x-0.3805) = \dfrac{3}{\sqrt{29}}, 0\le x\le 2\pi \end{gather*}$

Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.

Why do you set the domain for u between [0,pi]?

zsteve · « **Reply #70 on:** July 31, 2016, 08:57:52 pm »

I set $u\in[0, \pi/2]$ so that both sin and cos are positive, so I can use $\cos(x) = +\sqrt{1-\sin^2(x)}$ to avoid ambiguity regarding the sign of cos(x)

StupidProdigy · « **Reply #71 on:** July 31, 2016, 09:08:18 pm »

Quote from: RuiAce on July 31, 2016, 07:51:41 pm

Do they teach the tangent half-angle substitution or the auxiliary angle method in this course?

Nope! (at least not in units 3/4 specialist)

excelsiorxlcr · « **Reply #72 on:** August 04, 2016, 05:40:39 pm »

Please help me ASAP! I have a test very soon on this. A tram decelerates uniformly from a speed of 60 km/h to rest in 60s. Find the time taken for it to travel half the total distance. Thank you so much in advance

Sine · « **Reply #73 on:** August 05, 2016, 01:22:54 am »

Quote from: excelsiorxlcr on August 04, 2016, 05:40:39 pm

Please help me ASAP! I have a test very soon on this. A tram decelerates uniformly from a speed of 60 km/h to rest in 60s. Find the time taken for it to travel half the total distance. Thank you so much in advance

Use constant acceleration formulae

So to find the total displacement : s
$u = \frac{50}{3} (m/s)$
$t=60 (s)$
$v = 0 (m/s)$
$s = ?$
$s= 0.5 (u+v)t$
$s =0.5 (\frac{50}{3})(60)$
$s = 500 (m)$
so half the distance is 250m
and
$u = \frac{50}{3} (m/s)$
$v = 0 (m/s)$
$s = 500 (m)$
$a=?$
$v^2 =u^2 + 2as$
$0= (\frac{50}{3})^2+1000a$
$a=\frac{-5}{18} (m/s^2)$

now
$s=250(m)$
$u=\frac{50}{3} ( m/s)$
$a=\frac{-5}{18} (m/s^2)$
$t=?$
$s=ut+0.5at^2$
$250=\frac{50}{3}t+0.5\frac{-5}{18}t^2$
$t=17.6s$

anotherworld2b · « **Reply #74 on:** August 06, 2016, 12:38:32 am »

Quote from: RuiAce on July 24, 2016, 11:46:51 pm

Are these proofs?
$\begin{align*}LHS&=\sin 4\theta\\ &= \sin (2\times 2\theta)\\ &= 2\sin2\theta \cos 2\theta\\ &= 2(2\sin\theta\cos\theta)(\cos^2\theta - \sin^2\theta)\\ &= 4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta\\&=RHS\end{align*}$

I am still a bit confused did you use the double angle formula for the 3rd line? Also where did the cos^2theta - sin^theta come from in line 4?

How would you prove q19?

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