Damien_Dalgaard
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#1
if theta represents an angle such that sin2theta= tan theta - cos2theta,
then sintheta - costheta = ?

A - root2
B 0
C 1
D 2 root 2
E It cannot be determined

Yes you can put it into an equation solver and get the right answer.

But may I ask how you did it?
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ThatPerson
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(Original post by Damien_Dalgaard)
if theta represents an angle such that sin2theta= tan theta - cos2theta,
then sintheta - costheta = ?

A - root2
B 0
C 1
D 2 root 2
E It cannot be determined

Yes you can put it into an equation solver and get the right answer.

But may I ask how you did it?
Is the answer B?
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Damien_Dalgaard
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(Original post by ThatPerson)
Is the answer B?
I thought that too by Newton Raphson but the answer is apparently - root 2. Plugged it in to my casio fx991es not sure why though..
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ThatPerson
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(Original post by Damien_Dalgaard)
I thought that too by Newton Raphson but the answer is apparently - root 2. Plugged it in to my casio fx991es not sure why though..
Ah, I see it now. There is more than one solution because x can take multiple values.

Spoiler:
Show

 \sin 2x = \tan x - \cos 2x

After some rearranging and using  \cos 2x = 1- 2\sin^{2}x we arrive at


\dfrac{\sin x (2\cos^{2} x -1) + \cos x (1 - 2\sin^{2} x)}{\cos x}=0



\dfrac{\cos 2x(\sin x + \cos x)}{\cos x} = 0



\implies \cos 2x(\sin x + \cos x) = 0



\rightarrow (\cos x + \sin x)^{2}(\cos x - \sin x)=0



\implies x= \dfrac{3\pi}{4}
or  x = \dfrac{\pi}{4}

The latter results in 0, whilst the former produces  \sqrt{2} . Hence the solution to  \sin x - \cos x cannot be determined.
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Gawain
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(Original post by ThatPerson)
Ah, I see it now. There is more than one solution because x can take multiple values.

Spoiler:
Show

 \sin 2x = \tan x - \cos 2x

After some rearranging and using  \cos 2x = 1- 2\sin^{2}x we arrive at


\dfrac{\sin x (2\cos^{2} x -1) + \cos x (1 - 2\sin^{2} x)}{\cos x}=0



\dfrac{\cos 2x(\sin x + \cos x)}{\cos x} = 0



\implies \cos 2x(\sin x + \cos x) = 0



\rightarrow (\cos x + \sin x)^{2}(\cos x - \sin x)=0



\implies x= \dfrac{3\pi}{4}
or  x = \dfrac{\pi}{4}

The latter results in 0, whilst the former produces  \sqrt{2} . Hence the solution to  \sin x - \cos x cannot be determined.
I did it in a different way but got the same answers, not sure why sin(x)-cos(x) is undetermined.

sin(2x)+cos(2x)=\frac{sin(x)}{co  s(x)}

Let sin(x)=s and cos(x)=c

Then

2sc+1-2s^2=\frac{s}{c}

2sc(c-s)=s-c

So 2sc=sin(2x)=-1 meaning x=n\pi - \frac{\pi}{4}

And if sin(x)-cos(x)=0 then x=n \pi - \frac{3 \pi}{4}.

For 0\leq x \leq 2\pi then sin(x)-cos(x)=0, -\sqrt{2}.

The answer is A then.
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ThatPerson
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(Original post by Gawain)
I did it in a different way but got the same answers, not sure why sin(x)-cos(x) is undetermined.

sin(2x)+cos(2x)=\frac{sin(x)}{co  s(x)}

Let sin(x)=s and cos(x)=c

Then

2sc+1-2s^2=\frac{s}{c}

2sc(c-s)=s-c

So 2sc=sin(2x)=-1 meaning x=n\pi - \frac{\pi}{4}

For 0\leq x \leq 2\pi then sin(x)-cos(x)=\pm \sqrt{2}.

The answer is A then.
Your answer also shows that it is undetermined. If there are two possible answers then you can't say the solution to (sinx - cosx) is either one, despite one of the possible answers being an option.
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Gawain
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(Original post by ThatPerson)
Your answer also shows that it is undetermined. If there are two possible answers then you can't say the solution to (sinx - cosx) is either one, despite one of the possible answers being an option.
Since A and B are both true I guess it has to be the last one, but the answer didn't say 'undetermined' (i.e. needs a suitable domain for a singular answer) but rather 'cannot be determined'.

I took issue with the word 'cannot' because it clearly can be determined if given a suitable domain.
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ThatPerson
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(Original post by Gawain)
Since A and B are both true I guess it has to be the last one, but the answer didn't say 'undetermined' (i.e. needs a suitable domain for a singular answer) but rather 'cannot be determined'.

I took issue with the word 'cannot' because it clearly can be determined if given a suitable domain.
I view the two phrases as synonymous, and I think it's implied in both that the answer cannot be determined in an unrestricted domain.

I think this is more an issue of semantics rather than the actual answer, and additionally the last option also uses the phrase "cannot be determined".
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Damien_Dalgaard
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#9
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(Original post by ThatPerson)
Ah, I see it now. There is more than one solution because x can take multiple values.

Spoiler:
Show

 \sin 2x = \tan x - \cos 2x

After some rearranging and using  \cos 2x = 1- 2\sin^{2}x we arrive at


\dfrac{\sin x (2\cos^{2} x -1) + \cos x (1 - 2\sin^{2} x)}{\cos x}=0



\dfrac{\cos 2x(\sin x + \cos x)}{\cos x} = 0



\implies \cos 2x(\sin x + \cos x) = 0



\rightarrow (\cos x + \sin x)^{2}(\cos x - \sin x)=0



\implies x= \dfrac{3\pi}{4}
or  x = \dfrac{\pi}{4}

The latter results in 0, whilst the former produces  \sqrt{2} . Hence the solution to  \sin x - \cos x cannot be determined.
(Original post by Gawain)
I did it in a different way but got the same answers, not sure why sin(x)-cos(x) is undetermined.

sin(2x)+cos(2x)=\frac{sin(x)}{co  s(x)}

Let sin(x)=s and cos(x)=c

Then

2sc+1-2s^2=\frac{s}{c}

2sc(c-s)=s-c

So 2sc=sin(2x)=-1 meaning x=n\pi - \frac{\pi}{4}

And if sin(x)-cos(x)=0 then x=n \pi - \frac{3 \pi}{4}.

For 0\leq x \leq 2\pi then sin(x)-cos(x)=0, -\sqrt{2}.

The answer is A then.
Good methods, I kind of made a mess in doing in similar to @Gawain
Probably should have got that though.

So have you guys unanimously decided that its E.

With respect to the unrestricted domain, I suppose that would be considered too. I was was just quoting it as it was written?
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ThatPerson
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(Original post by Damien_Dalgaard)
Good methods, I kind of made a mess in doing in similar to @Gawain
Probably should have got that though.

So have you guys unanimously decided that its E.

With respect to the unrestricted domain, I suppose that would be considered too. I was was just quoting it as it was written?
I'm confident it's E.
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cxs
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Looking at this question, I think the quickest way is that, if you are familar with the transformation of tan(x/2), the relationship is such that sin2x=2t/(1+t2),cos2x=(1-t2)/(1+t2), and the question reduces to (t-1)(t+1)2=0. Where t is tanx.
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