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Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164278 times)  Share 

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9375 on: January 21, 2019, 06:51:02 pm »
0
If sin(x) = -3/4 and x is in the 3rd quadrant, what's sin(x/2)?
I tried subbing in a variable like alpha so
sin(2a)=2cos(a)sin(a)
sin(x)=2cos(x/2)sin(x/2)
Then solving for sin(x/2) but I'm still stuck with the cos part.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9376 on: January 21, 2019, 06:58:36 pm »
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1 word: Pythagorean (theorem.) hopefully that does the trick if not let me know
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9377 on: January 21, 2019, 09:33:16 pm »
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1 word: Pythagorean (theorem.) hopefully that does the trick if not let me know
still lost haha
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9378 on: January 21, 2019, 10:17:08 pm »
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still lost haha


Lol i just saw why. Oops. Im not really aure how you would approach this. I thought you were given a value for sin x/2 but you re not at all. And there is not really anyway I can think of taking care of that cos x/2. if it was squared we could do something pretty easy, but if its not, im not really sure. Sorry. I will keep trying.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9379 on: January 21, 2019, 11:05:56 pm »
+3
If sin(x) = -3/4 and x is in the 3rd quadrant, what's sin(x/2)?
I tried subbing in a variable like alpha so
sin(2a)=2cos(a)sin(a)
sin(x)=2cos(x/2)sin(x/2)
Then solving for sin(x/2) but I'm still stuck with the cos part.

The cosine double angle formula is often more versatile, as it involves only one trig function per side.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9380 on: January 22, 2019, 01:22:09 am »
0
If sin(x) = -3/4 and x is in the 3rd quadrant, what's sin(x/2)?
I tried subbing in a variable like alpha so
sin(2a)=2cos(a)sin(a)
sin(x)=2cos(x/2)sin(x/2)
Then solving for sin(x/2) but I'm still stuck with the cos part.
You don't HAVE to use the cosine double angle formula, but I do indeed suggest it, as mzhao points out. Otherwise, you're left with working out like this.
\(
\frac{9}{16} = \sin^2(x) = 4\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\\
= 4\sin^2\left(\frac{x}{2}\right)\left(1 - \sin^2\left(\frac{x}{2}\right)\right)\\
\sin^4\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)  + \frac{9}{64} = 0\\
\sin^2\left(\frac{x}{2}\right) = \frac{1\pm\sqrt{1 - \frac{9}{16}}}{2} = \frac{4\pm\sqrt{7}}{8}
\)
Now, we have a seriously awkward dilemma, and this is why you don't go around squaring things (you introduce extra solutions), as both signs admit valid solutions. As a heuristic, if \(\sin(x) = -0.75\), it is around \(-0.866\), so \(x\approx \frac{4\pi}{3}, \frac{x}{2}\approx \frac{2\pi}{3}\). As such, we expect \(\sin^2\left(\frac{x}{2}\right) \approx \frac{3}{4}\), which warrants taking the positive root as the negative root gives a number closer to 0.5. Indeed, it can be shown that \(\cos^2\left(\frac{x}{2}\right) = \frac{4 - \sqrt{7}}{8}\). Anyway, continuing on and accounting for the quadrant of \(\frac{x}{2}\) gives the same result as mzhao.

Edit: TeX on AN is flat-out not working properly for me for some reason. I had to run the code differently to make it show at all.
« Last Edit: January 22, 2019, 08:28:34 pm by lzxnl »
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9381 on: January 22, 2019, 04:26:03 pm »
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Thanks mzhao and lzxnl. Just one more question; when there's a dilation from the y axis ex. sin(x/2), sin(2x) etc. and the question states that x is in a specific quadrant, how do you know when the flip the negative to a positive or vice versa? Like I understand how sin(x), cosec(x), cot(x) are all positive and negative in certain quadrants, but don't get how dilations affect that.
« Last Edit: January 22, 2019, 04:28:01 pm by undefined »
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9382 on: January 22, 2019, 04:48:46 pm »
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Thanks mzhao and lzxnl. Just one more question; when there's a dilation from the y axis ex. sin(x/2), sin(2x) etc. and the question states that x is in a specific quadrant, how do you know when the flip the negative to a positive or vice versa? Like I understand how sin(x), cosec(x), cot(x) are all positive and negative in certain quadrants, but don't get how dilations affect that.

Okay hopefully I can answer this one.  :-[

So if you are given a domain of 0 to 2pi, for x, then you know the domain for 2x is going to be 0 to 4pi. In this way, there can be solutions in one or multiple quadrants. Hope this helps. Sorry for confusing you, my explananation here is very poor. 
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9383 on: January 22, 2019, 09:09:13 pm »
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Okay hopefully I can answer this one.  :-[

So if you are given a domain of 0 to 2pi, for x, then you know the domain for 2x is going to be 0 to 4pi. In this way, there can be solutions in one or multiple quadrants. Hope this helps. Sorry for confusing you, my explananation here is very poor. 
Oh I see, so for dilations like x/2 and a restricted domain of, say, the first quadrant, the solution has to be positive regardless since 0 to pi/4 would still be in the first quadrant?
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9384 on: January 22, 2019, 09:33:17 pm »
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Oh I see, so for dilations like x/2 and a restricted domain of, say, the first quadrant, the solution has to be positive regardless since 0 to pi/4 would still be in the first quadrant?


Yep.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9385 on: January 24, 2019, 04:28:52 pm »
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Why is the domain of arccos(x^2) [-1,1]?
I tried to solve -1<=x^2<=1 to find the new domain so shouldn't all negative numbers not count as they can't be square rooted? Thanks
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9386 on: January 24, 2019, 04:32:54 pm »
+2
Why is the domain of arccos(x^2) [-1,1]?
I tried to solve -1<=x^2<=1 to find the new domain so shouldn't all negative numbers not count as they can't be square rooted? Thanks
Edit: Misread.

Keep in mind that \(-1 \leq x^2 \leq 1\) is a restriction on the square of \(x\), not the square root. If you start with, say \(x=-\frac12\), you'll obtain \(x^2 = \frac14\) which is fine.
« Last Edit: January 24, 2019, 04:36:17 pm by RuiAce »

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9387 on: January 24, 2019, 05:04:04 pm »
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Edit: Misread.

Keep in mind that \(-1 \leq x^2 \leq 1\) is a restriction on the square of \(x\), not the square root. If you start with, say \(x=-\frac12\), you'll obtain \(x^2 = \frac14\) which is fine.
I'm confused, is -1<=x^2<=1 even valid in this case? as x^2=-1 when solving for x does not exist. Generally how I would go about finding domains is by isolating the 'x' so like for arcsin(3x+2) I would solve -1<=3x+2<=1 whereas for arccos(x^2), should you go from -1<=x<=1 and then square everything to see if the domain holds? i.e. -1<=x<=1 therefore 0<=x^2<=1 which holds for arccos(x^2). Not sure what the working out process is when there is a variable to a degree inside. Thanks for the reply by the way
« Last Edit: January 24, 2019, 05:07:29 pm by undefined »
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9388 on: January 24, 2019, 05:10:09 pm »
+2
Domain of arcos is 1 to -1 because for cos to be one to one, we need to restrict its domain to 0 to pi. Then, the range of cos is 1 to -1 and so the domain for the inverse cos function is -1 to 1.

If u are trying to find domain of the function arccos 3x +2, follow these steps

domain of x is -1 to 1.
Domain of 3x is -3 to 3.
Domain of 3x + 2 is -1 to 5.

Alternatively, find the inverse of the arccos function, so the cos function, and then find the range of that. 

Hope this helps. If I have done something wrong, or you dont understand please let me know.  :)
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9389 on: January 24, 2019, 05:10:19 pm »
+3
I'm confused, is -1<=x^2<=1 even valid in this case? as x^2=-1 does not exist. Generally how I would go about finding domainds is by isolating the 'x' so like for arcsin(3x+2) I would solve -1<=3x+2<=1 whereas for arccos(x^2), should you go from -1<=x<=1 and then square everything to see if the domain holds? i.e. -1<=x<=1 therefore 0<=x^2<=1 which holds for arccos(x^2). Not sure what the working out process is when there is a variable to a degree inside. Thanks for the reply by the way
\[ \text{Finding domains of }\arcsin[f(x)]\text{ and }\arccos[f(x)]\text{ always boils down somehow}\\ \text{to solving the equation }-1\leq f(x)\leq 1. \]
\[ \text{So your initial conclusion of }-1\leq x^2\leq 1\text{ is correct}\\ \text{and you're right to notice that }x^2=-1\text{ has no solution, for example.} \]
\[ \text{The idea is that because no real }x\text{ will satisfy }-1\leq x^2 < 0,\\ \text{this is nothing but a redundancy in our equation.}\\ \text{Our equation effectively boils down to solving }\boxed{0\leq x^2\leq 1}\text{ (i.e. that equation you mentioned)}\\ \text{which you can then check has the solution }-1\leq x \leq 1\text{ as you required.}\]
Solving inequalities is usually done on a case by case basis. However one nice universal approach (for non-linear inequalities) is just to treat them like an equation, but then sketch out some graph to determine which intervals are required in your inequality's solution.
« Last Edit: January 24, 2019, 05:12:06 pm by RuiAce »