VCE Specialist 3/4 Question Thread!

[^^^111^^^]

Hi AN,

I have a doubt..
Does anyone know, how to find the value of this expression by hand if it was cos(3cos^-1(1/a))?

So what I tried was using the double angle formula for cos(2x) = 2cos^2(x) -1 as I substituted the arccos for A and then simplified it... but the thing is I'm not sure if this will work if there was something different inside the cos, like cos(3A) or cos(4A) ,(where "A" would be the arccos(1/a)).

If asked, is it possible to do such a question by hand instead of going through the long process of substitution? And if so, please explain how.


Thanks!

[caffinatedloz]

Hi AN,

I have a doubt..
Does anyone know, how to find the value of this expression by hand if it was cos(3cos^-1(1/a))?

So what I tried was using the double angle formula for cos(2x) = 2cos^2(x) -1 as I substituted the arccos for A and then simplified it... but the thing is I'm not sure if this will work if there was something different inside the cos, like cos(3A) or cos(4A) ,(where "A" would be the arccos(1/a)).

If asked, is it possible to do such a question by hand instead of going through the long process of substitution? And if so, please explain how.


Thanks!

Hi ^^^111^^^. Hopefully someone with more maths expertise can also jump in, but I would have used the same double angle formula as you did to solve the equation.

For even multiples of x (eg: cos(4(cos^-1(1/a))), cos(6(cos^-1(1/a)))) I would use the same double angle formula. So for cos(4(cos^-1(1/a))) I would make it cos(2A) by substituting A for 2(cos^-1(1/a))). For cos(6(cos^-1(1/a))) I would use the substyitution 3(cos^-1(1/a))).

For odd numbers I would first use cos(a+b), then a double angle formula where necessary. For cos(3x), make it = cos(x)cos(2x) – sin(x)sin(2x). Then use double angle formulas. (As seen in my VERY rough working).

I would solve this by substitution, even though it is a long process.

[^^^111^^^]

Can someone help (or hint) me how to approach this question? I tried various things with it such as sin(-arccos(x)) = sin(-sin(x)+pi/2)) and then using compound angle formula (letting A = arcsin(x)) or taking the negative out, but I'm not able to show that it is equal to -sqrt(1-x^2).

Preferably, it would be great if someone hints me in the right direction (e.g. like suggesting another approach or a formula for e.g.), but I don't mind seeing the full solution for this 'show' question.

Thanks!

[fun_jirachi]

If \(\sin \theta = x\), then \(\cos \theta = \sqrt{1-x^2}\); this comes as a result either the identity \(\sin^2 \theta + \cos^2 \theta = 1\) or  the right angled triangle you would have seen in previous years (probably the first bit of trig with the SOHCAHTOA)

That means we can say \(\cos (-\arccos x) = \sqrt{1-(\sin (-\arccos x))^2}\). Pretty sure you can work it out from here; the only possible stumbling block is correctly assessing why we take the negative root (think about the domain for sine when looking at this).

imo try and avoid compound angles where possible. Hope this helps

[^^^111^^^]

If \(\sin \theta = x\), then \(\cos \theta = \sqrt{1-x^2}\); this comes as a result either the identity \(\sin^2 \theta + \cos^2 \theta = 1\) or  the right angled triangle you would have seen in previous years (probably the first bit of trig with the SOHCAHTOA)

That means we can say \(\cos (-\arccos x) = \sqrt{1-(\sin (-\arccos x))^2}\). Pretty sure you can work it out from here; the only possible stumbling block is correctly assessing why we take the negative root (think about the domain for sine when looking at this).

imo try and avoid compound angles where possible. Hope this helps

Thanks that makes a lot of sense now!! 

[bw304]

hi all, im really stuck on this question from my holiday homework:

A yacht sails from A to B on a bearing of 075* for 5km then from B to C on a bearing of 315* for 10km. Find:

a) the exact distance between A and C
b) the exact distance that the yacht is from A when it is closest to A on the BC leg.

for part a) i got 5√3 km (i don't have access to solutions so not sure if this is correct)
and I'm really stuck on part b). like how do you find out which point of BC is closest to A??

any help/hints would be appreciated !!!

[caffinatedloz]

like how do you find out which point of BC is closest to A??

any help/hints would be appreciated !!!

To find the point that is closest you'll need to use the distance formula. Find the distance between A and a point (x,y) where (x,y) lies on the line BC. (Do this by using the information about BC in the question to write y in terms of x).

Once you have the distance between the points in terms of x, find the derivative. Then set the derivative equal to 0 and solve for x. This will give you maximum and minimum points. Test the x value(s) and find the minimum x value. Then plug it back into your formula for BC to find the point.

If you have any more questions or want me to have a go at the working if you're still stuck just let me know!

[bw304]

To find the point that is closest you'll need to use the distance formula. Find the distance between A and a point (x,y) where (x,y) lies on the line BC. (Do this by using the information about BC in the question to write y in terms of x).

Once you have the distance between the points in terms of x, find the derivative. Then set the derivative equal to 0 and solve for x. This will give you maximum and minimum points. Test the x value(s) and find the minimum x value. Then plug it back into your formula for BC to find the point.

If you have any more questions or want me to have a go at the working if you're still stuck just let me know!

oh yes that makes sense! thanks so much

[galaxysauce]

Hi! Could someone please answer this question for 3bii.. Thanks! 

[fun_jirachi]

Given two complex numbers \(u\) and \(v\), the midpoint of the two is always represented by \(\frac{u+v}{2}\).

What can you then deduce about \(M\) in this question?

[galaxysauce]

I can deduce that M is the midpoint. I got the points and found the midpoint of it to find M.. I worked it out last night but my brain just wasn't working lol. Thanks for the help! I'll come back with more questions in the future 

[galaxysauce]

Hi.. Could someone please explain this question? Thanks!

[^^^111^^^]

Hi.. Could someone please explain this question? Thanks!
(Image removed from quote.)

Hi,

Hopefully someone can correct me if I'm not right, but I think the question is asking you to simplify the expressions and rewrite in the form z = r cis(θ) .

For example, for part a), try to think of how you would write the conjugate of a complex number, and hence z = cos(θ) - i sin(θ) in the form z = r cis(θ), and work your way from that.

[galaxysauce]

Hi.. I'm back again haha. 
Could someone please help me with this question? P.S I've noticed that I have been struggling with proof questions and was wondering if anyone knew how to get better at them? Thanks!

[ArtyDreams]

Hi.. I'm back again haha. 
Could someone please help me with this question? P.S I've noticed that I have been struggling with proof questions and was wondering if anyone knew how to get better at them? Thanks!
(Image removed from quote.)

Hi!

For this question, the main point is that ABX is an equilateral triangle, therefore, all angles and sides are equal. The angles in an equilateral triangle are 60º, so you use angle properties to deduce the sizes of the other angles. (For example, angle XAD is 30º, since XAB is 60º, and they add up to a right angle).

Hope this helps, let me know if you need some more guidance!