YEAR 11 TRIGONOMETRY EXAM (1 Viewer)

od9895 · Jan 11, 2012

Hello
I am about to do a year 11 trigonometry exam at truongs in a few weeks time and I was wondering if any of you guys who gone to truongs annd sat for that trig exam have an tips or hints about the exam and what type of questions or specific things i have to study.

Please let me know

Thanks

Carrotsticks · Jan 11, 2012

Study trigonometry.

cutemouse · Jan 11, 2012

od9895 said:
Hello
I am about to do a year 11 trigonometry exam at truongs in a few weeks time and I was wondering if any of you guys who gone to truongs annd sat for that trig exam have an tips or hints about the exam and what type of questions or specific things i have to study.

Please let me know

Thanks

Try this:

$\text{Solve for } 0 \le \theta \le 360^{\circ}: \ \sin 2 \theta = \frac{1}{\sqrt{2}}$

hayabusaboston · Jan 11, 2012

^ is that yr 11 ext maths? thought it was in advanced..?

Carrotsticks · Jan 11, 2012

hayabusaboston said:
^ is that yr 11 ext maths? thought it was in advanced..?

Nope. Even though it is $sin2\theta$ , double angle formulae are not required to do the question.

hayabusaboston · Jan 11, 2012

Carrotsticks said:
Nope. Even though it is $sin2\theta$ , double angle formulae are not required to do the question.

Yea I mean I thought I read above "yr 11 ext 1 maths exam". Didn't read properly lol

cutemouse · Jan 11, 2012

Carrotsticks said:
Nope. Even though it is $sin2\theta$ , double angle formulae are not required to do the question.

In fact you can't do it using the double angle formula...

lolcakes52 · Jan 12, 2012

You could but it would be annoying, much easier to do it the other way. Do some 3-d trig stuff and look at past papers.

Carrotsticks · Jan 12, 2012

cutemouse said:
In fact you can't do it using the double angle formula...

Actually, you can.

$\begin{align*}\sin 2 \theta &= \frac{1}{\sqrt{2}} \\ 2\sin \theta \cos \theta&= \frac{1}{\sqrt{2}} \\ 2\sin \theta &= \frac{1}{\sqrt{2}\cos \theta} \end{align*} \\\\ $Divide both sides by $ \cos \theta: \\ \begin{align*}2\tan \theta &= \frac{1}{{\sqrt{2}}\cos^{2} \theta} \\ &= \frac{1}{\sqrt{2}} \sec^{2} \theta \\ &= \frac{1}{\sqrt{2}} (1+\tan^{2} \theta) \\ \end{align*} \\ $Re-arrange as a quadratic in terms of $ \tan \theta: \\\\ \tan^{2}\theta - 2\sqrt{2} \tan \theta + 1 =0 \\\\ $Using the quadratic formula, we acquire $\tan \theta = \sqrt{2} \pm {1} \\\\ $We know that $ \tan^{-1} (\sqrt{2} -1) $ is $ \frac{\pi}{8} $ and $ \tan^{-1} (\sqrt{2} +1) $ is $ \frac{3\pi}{8} \\\\ $Using the 'cousin angles', we also acquire the solutions $ \frac{9\pi}{8} $ and $ \frac{11\pi}{8}$

hayabusaboston · Jan 12, 2012

Carrotsticks said:
Actually, you can.

$\sin 2 \theta = \frac{1}{\sqrt{2}} \\ 2\sin \theta \cos \theta \frac{1}{\sqrt{2}} \\\\ 2\sin \theta = \frac{1}{\sqrt{2}\cos \theta} \\\\ $Divide both sides by $ \cos \theta \\\\ 2\tan \theta = \frac{1}{{\sqrt{2}}\cos^{2} \theta}$

...

theta=22.5,67.5,202.5,247.5

Drongoski · Jan 12, 2012

Carrotsticks said:
Nope. Even though it is $sin2\theta$ , double angle formulae are not required to do the question.

Right.

$sin 2 \theta = \frac {1}{\sqrt {2}} \implies 2 \theta = 45 ^ \circ \implies \theta = \frac {45 ^ \circ}{2}$

hayabusaboston · Jan 12, 2012

Drongoski said:
Right.

$sin 2 \theta = \frac {1}{\sqrt {2}} \implies 2 \theta = 45 ^ \circ \implies \theta = \frac {45 ^ \circ}{2}$

You gotta get the first two quadrants, so 45 and 135, then plus 360 because of 2 theta, then divide everything by 2. So as my last post.

Carrotsticks · Jan 12, 2012

hayabusaboston said:
You gotta get the first two quadrants, so 45 and 135, then plus 360 because of 2 theta, then divide everything by 2. So as my last post.

I'm pretty sure he's already aware of this, but just couldn't be bothered typing it down.

LaTeX can be a pain sometimes.

RivalryofTroll · Jan 12, 2012

cutemouse said:
Try this:

$\text{Solve for } 0 \le \theta \le 360^{\circ}: \ \sin 2 \theta = \frac{1}{\sqrt{2}}$

ASTC questions are always easy marks!

@OP:
3D - trig (elevation) questions are what you got to watch out for! (if you even did it)
And according to my friend, some truong classes have already sat that trig-exam that you are about to do so if you are desperate for help, you know what to do (ask them about the test).
Also out of curiosity, what class are you in? I heard truong classes are divided into school groups like there's a Hurlstone Class, Ruse Class, Macquarie Fields Class, Sefton Class, etc.

hit patel · Apr 13, 2013

Or u can do this:
Sin2theta=1/sqrt(2)

u=2theta

Sinu=1/sqrt(2) (0≤x≤720)

u=45, 135, 405, 495
Since u=2theta
Theta=u/2
therefore theta= 45/2, 135/2, 405/2, 495/2 degrees.

Much simpler method than the above ones.

melkhoury2145 · Apr 26, 2016

cutemouse said:
Try this:

$\text{Solve for } 0 \le \theta \le 360^{\circ}: \ \sin 2 \theta = \frac{1}{\sqrt{2}}$

Since it is 1/root 2 it can have positive and negative solutions. Therefore having 8 solutions in total where theta is equal to +- 22.5, 67.5, 112.5, 157.5, 202.5, 247.5, 292.5, 337.5

si2136 · Apr 26, 2016

melkhoury2145 said:
Since it is 1/root 2 it can have positive and negative solutions. Therefore having 8 solutions in total where theta is equal to +- 22.5, 67.5, 112.5, 157.5, 202.5, 247.5, 292.5, 337.5

Please try not to bump old threads.

Green Yoda · Apr 26, 2016

cutemouse said:
Try this:

$\text{Solve for } 0 \le \theta \le 360^{\circ}: \ \sin 2 \theta = \frac{1}{\sqrt{2}}$

Note: 0<=2θ<=720
Therefore 2θ = 45°,135°,405°,495°
Therefore θ = 22.5°,67.5°,202.5°,247.5°

Green Yoda · Apr 26, 2016

melkhoury2145 said:
Since it is 1/root 2 it can have positive and negative solutions. Therefore having 8 solutions in total where theta is equal to +- 22.5, 67.5, 112.5, 157.5, 202.5, 247.5, 292.5, 337.5

It cannot have neg solutions..

eyeseeyou · Apr 29, 2016

wtf this thread is pree old

YEAR 11 TRIGONOMETRY EXAM (1 Viewer)

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