Specialist 1/2 Question Thread!

[keltingmeith]

Unfortunately I don't have answers; it's an old school 1/2 Specialist exam, but the answers are just missing for this section A.
If we let a=65, I got the same answers as you.
Side note, how does your 'alternatively' section work? As in how can we say a+c+90=180? Is it still to do with the opposite angles of cyclic quadrilateral sum to 180?

That's using the theorem that an angle subtended on a circle from the diameter is always 90 degrees

[redset8]

That's using the theorem that an angle subtended on a circle from the diameter is always 90 degrees

Ah yeah now I see it.
Thanks heaps for your help!!!

[xpikachuzz]

this one was one my year 11 exam
if i have the values of sin(x) and cos(x) then what is sin(x/2).
im guesing i need to expand it but i dont know how to... on my exam i guessed sin(x/2) = sin(x)cos(x)

[keltingmeith]

this one was one my year 11 exam
if i have the values of sin(x) and cos(x) then what is sin(x/2).
im guesing i need to expand it but i dont know how to... on my exam i guessed sin(x/2) = sin(x)cos(x)

Yeah, not quite! You've mixed up the double-angle formula \(\sin(2u)=2\sin(u)\cos(u)\). I could re-phrase this using u=x/2, and get:

\[
\sin(x)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)
\]

Which, as you can see, is very different to what you did! You'd need to be looking into some other double angle formula - but here's a hint. If you know that \(\sin(\alpha)=\beta\), then you also know that \(\cos(\alpha)=\sqrt{1-\beta^2}\). You can also do this in reverse, and if \(\cos(\alpha)=\beta\), then \(\sin(\alpha)=\sqrt{1-\beta^2}\) You may not need to use my hint, either. There is a method that doesn't need that hint, but hopefully it's stimulating some thought

[Ruchir]

Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks

[Bri MT]

Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks

Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:

This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)

Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n

Let's put this together:

a_n is the sum of (-1)^n x^2n

we are told that m is even (so that gives us a plus for our last term)


Hope this helps!

[Ruchir]

Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:

This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)

Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n

Let's put this together:

a_n is the sum of (-1)^n x^2n

we are told that m is even (so that gives us a plus for our last term)


Hope this helps!

I appreciate the effort you took to solve this question but the answer is a bit different
This is the answer is a bit different.

[keltingmeith]

Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks

Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine

[Ruchir]

Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine

Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms

[keltingmeith]

Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms

Okay, well try breaking the series up and see if that makes things easier. Eg,
When m=1, you go up to x^(2m)=x^2: 1 - x^2
When m=4, you go up to x^(2m)=x^8: 1 - x^2 + x^4 - x^6 + x^8

Try a few more, fill out the series, count up the terms, and see what you notice

[Ruchir]

Okay, well try breaking the series up and see if that makes things easier. Eg,
When m=1, you go up to x^(2m)=x^2: 1 - x^2
When m=4, you go up to x^(2m)=x^8: 1 - x^2 + x^4 - x^6 + x^8

Try a few more, fill out the series, count up the terms, and see what you notice

Thanks I got it.
Just 1 quick question
When they say m is even, do they mean all values of m are supposed to be even?

[Ruchir]

I don’t understand when for 9 a and b.
Here’s what I did
for A. 6*6*5*3 = 540 Pr(even) = 540/1470
For B just Pr(odd) = 930/1470
But the answer if for B is 720/1470
And for A is 750/1470
Please help and explain what I did wrong

[cloudyy]

Could someone help me with this q?

From the top of a cliff 160m high, two buoys are observed. Their bearings are 337 and 308. Their respective angles of depression are 3 and 5. Calculate the distance between the buoys.

I've tried drawing a diagram to figure it out, but I'm way too confused on how to draw the diagram with bearing included. Any tips?

[fun_jirachi]

Hey there!

Open for a diagram

If you open the above spoiler, I've drawn a diagram on paint. Hopefully it's clear enough, the green triangle corresponds to the buoy with bearing 337 and the red triangle corresponds to the buoy with bearing 308.

Try using different colours and angle your North-South axis a bit so you leave space for the vertical axis for 3D diagrams like this, and feel free to exaggerate the angles a bit as diagrams don't really need to be accurate as they only serve as an interpretive tool.

Further help in the spoiler below if you need it, just in case of followups:

Spoiler

Find the distance from the cliff to each of the buoys using trig ratios, then use the cosine rule to find the distance between the buoys.

Hope this helps

[cloudyy]

Hey there!

Open for a diagram

If you open the above spoiler, I've drawn a diagram on paint. Hopefully it's clear enough, the green triangle corresponds to the buoy with bearing 337 and the red triangle corresponds to the buoy with bearing 308.

Try using different colours and angle your North-South axis a bit so you leave space for the vertical axis for 3D diagrams like this, and feel free to exaggerate the angles a bit as diagrams don't really need to be accurate as they only serve as an interpretive tool.

Further help in the spoiler below if you need it, just in case of followups:

Spoiler

Find the distance from the cliff to each of the buoys using trig ratios, then use the cosine rule to find the distance between the buoys.

Hope this helps

You're my lifesaver!! Can't thank you enough, thank you!