HSC 2016 MX1 Marathon (archive) (2 Viewers)

Drsoccerball · Feb 2, 2016

Re: Flop math question thread

Flop21 said:
What's the formula we're talking about here?

Its finding all values for a trig equation.

Eg: $\sin{(x)} = 0$

$x= \pm \pi k $ Where k is any integer $$

Flop21 · Feb 2, 2016

Re: Flop math question thread

Drsoccerball said:
Its finding all values for a trig equation.

Eg: $\sin{(x)} = 0$

$x= \pm \pi k $ Where k is any integer $$

Oh interesting I haven't seen that one before, thanks.

Flop21 · Feb 2, 2016

Re: Flop math question thread

cosine(x) = sin(x/2)

Solve the equation. It's an acute angle it tells you.

I'm stuck on this, as are some other people.

Drsoccerball · Feb 2, 2016

Re: Flop math question thread

Flop21 said:
Oh interesting I haven't seen that one before, thanks.

Okay so how you determine these is like this :

$$ If $ 0< \theta < \frac{\pi}{2}$

$\sin{x} = \theta$

$x= \sin ^{-1}(\theta) \pm 2 \pi k $ or (using ambiguity of sin in first and second quadrant ) $ x= \pi - \sin ^{-1}(\theta) \pm 2 \pi k$

InteGrand · Feb 2, 2016

Re: Flop math question thread

Flop21 said:
cosine(x) = sin(x/2)

Solve the equation. It's an acute angle it tells you.

I'm stuck on this, as are some other people.

$\noindent Note that $\sin \frac{x}{2} \geq 0$ as $x$ is acute. So we use this trig. identity (sub. this for the R.H.S. of the given equation): $\sin \frac{x}{2}= \sqrt{\frac{1}{2} -\frac{1}{2}\cos x}$ (taking non-negative square root as we know the sine is non-negative). Then we can square both sides of the to-solve equation $\cos x =\sqrt{\frac{1}{2}- \frac{1}{2}\cos x}$ obtain a quadratic in $\cos x$ to solve. Make sure to only take acute solutions, and test each solution to double-check, because squaring both sides of an equation we are trying to solve can introduce spurious solutions.$

$\noindent (Technically we won't need to worry about spurious solutions for this one, since by taking only acute solutions, we obtain solutions where the LHS and RHS of the equation we had before squaring both sides have the same sign.)$

InteGrand · Feb 2, 2016

Re: Flop math question thread

Drsoccerball said:
Its finding all values for a trig equation.

Eg: $\sin{(x)} = 0$

$x= \pm \pi k $ Where k is any integer $$

$\noindent By the way, we don't need to write the $\pm$, because integers can be negative as well, so saying that $k$ is any integer is already sufficient.$

leehuan · Feb 2, 2016

Re: Flop math question thread

Drsoccerball said:
Also when you're doing general solution make sure NOT to remember the formula and derive it yourself (takes 2 seconds).

Although she isn't an HSC student anymore obviously, I believe it is on the formula sheet for subsequent years.

But what's so hard about that? I never 'derived' a general equation formula.

sin(x)=a -> x=(-1)^n.arcsin(a) + n.pi
cos(x)=a -> x=2n.pi ± arccos(a)
tan(x)=a -> x=pi + arctan(a)

Flop21 · Feb 2, 2016

Re: Flop math question thread

InteGrand said:
$\noindent Note that $\sin \frac{x}{2} \geq 0$ as $x$ is acute. So we use this trig. identity (sub. this for the R.H.S. of the given equation): $\sin \frac{x}{2}= \sqrt{\frac{1}{2} -\frac{1}{2}\cos x}$ (taking non-negative square root as we know the sine is non-negative). Then we can square both sides of the original equation and obtain a quadratic in $\cos x$ to solve. Make sure to only take acute solutions, and test each solution, because squaring both sides of an equation we are trying to solve can introduce spurious solutions.$

Thanks I understand how to do it this way.

But how do we obtain that identity?

I'm guessing it's from this one, sin^2 (x) = 1/2 - 1/2 * cos(2x)

InteGrand · Feb 2, 2016

Re: Flop math question thread

leehuan said:
Although she isn't an HSC student anymore obviously, I believe it is on the formula sheet for subsequent years.

But what's so hard about that? I never 'derived' a general equation formula.

sin(x)=a -> x=(-1)^n.arcsin(a) + n.pi
cos(x)=a -> x=2n.pi ± arccos(a)
tan(x)=a -> x=pi + arctan(a)

Maybe he doesn't want people to memorise those.

leehuan · Feb 2, 2016

Re: Flop math question thread

Drsoccerball said:
Okay so how you determine these is like this :

$$ If $ 0< \theta < \frac{\pi}{2}$

$\sin{x} = \theta$

$x= \sin ^{-1}(\theta) \pm 2 \pi k $ or (using ambiguity of sin in first and second quadrant ) $ x= \pi - \sin ^{-1}(\theta) \pm 2 \pi k$

Why do this when the fact is when you take the arcsin of a negative you output a negative anyway?

arcsin(-x)=-arcsin(x)

InteGrand said:
Maybe he doesn't want people to memorise those.

That's... kind of insulting ._.

InteGrand · Feb 2, 2016

Re: Flop math question thread

Flop21 said:
Thanks I understand how to do it this way.

But how do we obtain that identity?

I'm guessing it's from this one, sin^2 (x) = 1/2 - 1/2 * cos(2x)

$\noindent Yeah, it's from that and replace $x$ with $\frac{x}{2}$. If you want to know how to obtain \textit{that} identity, it's in the HSC 3U Maths course and is a rearrangement of the double angle formula for cosine, which can be derived from a unit circle (see the Year 11 3 Unit Pender (Cambridge) textbook).$

Drsoccerball · Feb 2, 2016

Re: Flop math question thread

leehuan said:
Why do this when the fact is when you take the arcsin of a negative you output a negative anyway?

arcsin(-x)=-arcsin(x)

Why would I add another operation to the formula lol ?

That's... kind of insulting ._.

It's not something worthy of being memorised.

InteGrand · Feb 2, 2016

Re: Flop math question thread

leehuan said:
Why do this when the fact is when you take the arcsin of a negative you output a negative anyway?

That general solution formula is in the Year 12 3 Unit Pender textbook.

leehuan · Feb 2, 2016

Re: Flop math question thread

Drsoccerball said:
Why would I add another operation to the formula lol ?

It's not something worthy of being memorised.

You don't?

$\\ \sin{x} = -\frac{1}{\sqrt{2}} \\ x=n \pi + {(-1)}^{n} \frac{- \pi}{4}$

_______________________

Nor do you 'have' to derive it either.

InteGrand · Feb 2, 2016

Re: Flop math question thread

leehuan said:
You don't?

$\\ \sin{x} = -\frac{1}{\sqrt{2}} \\ x=n \pi \pm {(-1)}^{n} \frac{- \pi}{4}$

Maybe he prefers the other formula because it has more similarities to the ones for cos and tan?

leehuan · Feb 2, 2016

Re: Flop math question thread

InteGrand said:
Maybe he prefers the other formula because it has more similarities to the ones for cos and tan?

Meh. The ± is exactly how I remembered the cosine one. Because it's the only one that needed it to look more convenient.

InteGrand · Feb 2, 2016

Re: Flop math question thread

leehuan said:
Meh. The ± is exactly how I remembered the cosine one. Because it's the only one that needed it to look more convenient.

The one for the sine one that Drsoccerball wrote is more intuitive for most people, whereas the one you wrote is more elegant to write down (though some people find it harder to remember).

Flop21 · Feb 2, 2016

Re: Flop math question thread

InteGrand said:
$\noindent Note that $\sin \frac{x}{2} \geq 0$ as $x$ is acute. So we use this trig. identity (sub. this for the R.H.S. of the given equation): $\sin \frac{x}{2}= \sqrt{\frac{1}{2} -\frac{1}{2}\cos x}$ (taking non-negative square root as we know the sine is non-negative). Then we can square both sides of the to-solve equation $\cos x =\sqrt{\frac{1}{2}- \frac{1}{2}\cos x}$ obtain a quadratic in $\cos x$ to solve. Make sure to only take acute solutions, and test each solution to double-check, because squaring both sides of an equation we are trying to solve can introduce spurious solutions.$

I keep getting a power to 3, instead of 2 to solve for quadratic.

leehuan · Feb 2, 2016

Re: Flop math question thread

Lucky for any future HSC students they don't have to memorise them anymore. http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths-ref-sheet.pdf

Personally I actually found mine more intuitive because I thought ok wait a minute, 0≤arccos(x)≤π so by keeping a ± there with a 2π quantity, everything was preserved.

Whereas since -π/2≤arcsin(x)≤π/2 and similarly for arctan(x) without equality case, this range was far more suitable for just π to be present. Then leaving things as it is took care of tangent easily enough whereas the (-1)^n told me to in a way, go back and forth between 1st and 2nd for sine.

leehuan · Feb 2, 2016

Re: Flop math question thread

Try this Flop

$\\ \cos { 2x } =1-2\sin ^{ 2 }{ x } \\ \cos { x } =1-2\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } \\ \\ \cos { x } =\sin { \left( \frac { x }{ 2 } \right) } \\ 1-2\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } =\sin { \left( \frac { x }{ 2 } \right) }$

$\\ $Albeit, carefully noting that $ 0<x<\frac { \pi }{ 2 } \Rightarrow 0<\frac { x }{ 2 } <\frac { \pi }{ 4 }$

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