Trig Question (1 Viewer)

Ambility · Jan 28, 2016

$$Find the general solution to $\tan\theta =\sin 2\theta.$

My answer:

$\tan\theta = \sin 2\theta\\\frac{\sin\theta}{\cos\theta}=2\sin\theta\cos\theta\\\frac{1}{\cos\theta}=2\cos\theta\\1=2\cos^2\theta\\\cos\theta = \frac{1}{\pm \sqrt{2}}\\\theta = \frac{\pm n \pi}{4}$ where $n\in \mathbb{Z}$

The past paper answer sheet says:

$\theta = \pi n$ where $ n \in \mathbb{Z}$ or $\theta =\pi n\pm\frac{\pi}{4}$ where $n\in\mathbb{Z}$

What am I missing?

InteGrand · Jan 28, 2016

Ambility said:
$$Find the general solution to $\tan\theta =\sin 2\theta.$

My answer:

$\tan\theta = \sin 2\theta\\\frac{\sin\theta}{\cos\theta}=2\sin\theta\cos\theta\\\frac{1}{\cos\theta}=2\cos\theta\\1=2\cos^2\theta\\\cos\theta = \frac{1}{\pm \sqrt{2}}\\\theta = \frac{\pm n \pi}{4}$ where $n\in \mathbb{Z}$

The past paper answer sheet says:

$\theta = \pi n$ where $ n \in \mathbb{Z}$ or $\theta =\pi n\pm\frac{\pi}{4}$ where $n\in\mathbb{Z}$

What am I missing?

$\noindent When you divided out the $\sin \theta$ from both sides early on, you implicitly assumed that $\sin \theta \neq 0$ (since we can't divide through by 0). So you missed out the solutions for $\theta$ where $\sin \theta=0$. But clearly if $\sin \theta=0$, the original equation \textit{is} satisfied, since the L.H.S. is well-defined and 0 (as its denominator is non-zero whenever $\sin \theta =0$) and the R.H.S. is also 0, as it is $2\sin \theta \cos \theta$. So we also have as part of our solution the values of $\theta$ such that $\sin \theta = 0$, i.e. $\theta = n\pi,n\in \mathbb{Z}$ (since $\sin \theta = 0$ if and only if $\theta = n\pi$ for some $n\in \mathbb{Z}$).$

Ambility · Jan 28, 2016

InteGrand said:
$\noindent When you divided out the $\sin \theta$ from both sides early on, you implicitly assumed that $\sin \theta \neq 0$ (since we can't divide through by 0). So you missed out the solutions for $\theta$ where $\sin \theta=0$. But clearly if $\sin \theta=0$, the original equation \textit{is} satisfied, since the L.H.S. is well-defined and 0 (as its denominator is non-zero whenever $\sin \theta =0$) and the R.H.S. is also 0, as it is $2\sin \theta \cos \theta$. So we also have as part of our solution the values of $\theta$ such that $\sin \theta = 0$, i.e. $\theta = n\pi,n\in \mathbb{Z}$ (since $\sin \theta = 0$ if and only if $\theta = n\pi$ for some $n\in \mathbb{Z}$).$

In future, how can I avoid this?

InteGrand · Jan 28, 2016

Ambility said:
In future, how can I avoid this?

$\noindent When you come to something like $\sin \theta \times \text{bla}=\sin \theta \times\text{blah}$, instead of going straight to $\text{bla}=\text{blah}$, say something like ``hence $\sin \theta = 0$ or $\text{bla}=\text{blah}$''. (This is due to the fact that in general if $ab=ac$ where $a,b,c$ are real (or complex) numbers, then either $a=0$ or $b=c$, where ``or'' means the inclusive or, i.e. ``and/or''.)$

Ambility · Jan 28, 2016

InteGrand said:
$\noindent When you come to something like $\sin \theta \times \text{bla}=\sin \theta \times\text{blah}$, instead of going straight to $\text{bla}=\text{blah}$, say something like ``hence $\sin \theta = 0$ or $\text{bla}=\text{blah}$''. (This is due to the fact that in general if $ab=ac$ where $a,b,c$ are real (or complex) numbers, then either $a=0$ or $b=c$, where ``or'' means the inclusive or, i.e. ``and/or''.)$

Cheers mate.

leehuan · Jan 28, 2016

$\frac { \sin { x } }{ \cos { x } } =2\sin { x } \cos { x } \\ \sin { x } =2\sin { x } \cos ^{ 2 }{ x } \\ \sin { x } \left( 2\cos ^{ 2 }{ x } -1 \right) =0\\ \sin { x } \cos { \left( 2x \right) } =0\\ \sin { x } =0,\cos { 2x } =0\\ x=n\pi ,\quad 2x=2n\pi \pm \frac { \pi }{ 2 } \\ x=n\pi ,\quad x=n\pi \pm \frac { \pi }{ 4 }$

maintaining that n is an integer

Of course, IG has already explained the flaw in the cancelling out step. Refer to mine for alternate working.

Trig Question (1 Viewer)

Ambility

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InteGrand

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Ambility

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InteGrand

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Ambility

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leehuan

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