Trig. Help (1 Viewer)

Smile12345 · Aug 4, 2013

Hello All...

Could someone please see if I've gone through the right process for this question.

Q.
$$Solve for 0^{\circ} \leq \Theta \leq 360^{\circ}$
$2cos(2{\Theta}) = 1$$

So $cos(2{\Theta}) = 0.5$$

Then because we have 2(theta) in the middle, the range is between 0 and 180.

So the only answer would be cos(theta)= 60 degrees.

Thanks in advance.

Drongoski · Aug 4, 2013

It wants values of one theta between 0 and 180 deg.

2 angles (for 0 to 360)have cosine = 0.5, viz: 60 deg and 300 deg

$\therefore 2\theta = 60^{\circ} $ and $ 300^{\circ} \\ \\ \therefore \theta = 30^{\circ} $ and $ 150^{\circ}$

Smile12345 · Aug 4, 2013

Drongoski said:
It wants values of one theta between 0 and 180 deg.

2 angles (for 0 to 360)have cosine = 0.5, viz: 60 deg and 300 deg

$\therefore 2\theta = 60^{\circ} $ and $ 300^{\circ} \\ \\ \therefore \theta = 30^{\circ} $ and $ 150^{\circ}$

Yes, I understand what you've said. Thanks heaps...

Smile12345 · Aug 4, 2013

How about this one...

$$Solve for -180^{\circ} \leq \Theta \leq 180^{\circ}$
$2sin^2(3{\Theta}) = 1$$

Thanks.

Drongoski · Aug 4, 2013

Smile12345 said:
How about this one...

$$Solve for -180^{\circ} \leq \Theta \leq 180^{\circ}$
$2sin^2(3{\Theta}) = 1$$

Thanks.

$\therefore sin 3 \theta = \pm \frac {1}{\sqrt {2}}$

$\therefore $ over the reqd range: $ 3\theta = \pm 45^{\circ}, \pm 135^{\circ}, \pm 225^{\circ}, \pm 315^{\circ}, \pm 405^{\circ}, \pm 495^{\circ}$

Divide by 3 for single theta.

cookiez69 · Aug 4, 2013

Drongoski said:
It wants values of one theta between 0 and 180 deg.

2 angles (for 0 to 360)have cosine = 0.5, viz: 60 deg and 300 deg

$\therefore 2\theta = 60^{\circ} $ and $ 300^{\circ} \\ \\ \therefore \theta = 30^{\circ} $ and $ 150^{\circ}$

The domain is $0^{\circ} \leq \theta \leq 360^{\circ}$

So its not values of one theta between 0 and 180 degrees, but one theta between 0 and 360 as the question states. What you should be getting is

$2\theta = 60, 300, 420, 660 \\ \therefore \theta = 30, 150, 210, 330$

pdang · Aug 4, 2013

if the range for theta os 0 to 360, then the range for 2theta is 0 to 720
you should get 4 answers

pdang · Aug 4, 2013

lol beat me

Smile12345 · Aug 4, 2013

Drongoski said:
$\therefore sin 3 \theta = \pm \frac {1}{\sqrt 2}$

And $\pm \frac {1}{\sqrt 2}$ = Sin45....

Smile12345 · Aug 4, 2013

Drongoski said:
$\therefore sin 3 \theta = \pm \frac {1}{\sqrt {2}}$

$\therefore $ over the reqd range: $ 3\theta = \pm 45^{\circ}, \pm 135^{\circ}, \pm 225^{\circ}, \pm 315^{\circ}, \pm 405^{\circ}, \pm 495^{\circ}$

Divide by 3 for single theta.

Ok, understand you... Lucky it isn't a number like 8!!

Thanks.

+1

Drongoski · Aug 4, 2013

cookiez69 said:
The domain is $0^{\circ} \leq \theta \leq 360^{\circ}$

So its not values of one theta between 0 and 180 degrees, but one theta between 0 and 360 as the question states. What you should be getting is

$2\theta = 60, 300, 420, 660 \\ \therefore \theta = 30, 150, 210, 330$

When I did the question, I thought I saw the the range for theta from 0 to 180. Hence wording of my first line.

Thanks to you & pdang for pointing out my error. Goes to show I'm not infallible after all.

Smile12345 · Aug 4, 2013

So for a more simple one like

$$Solve for 0^{\circ} \leq \Theta \leq 360^{\circ}$
$sin({\Theta}) = - Root 3 over 2$$

Would there just be two answers... 240 and 360?

Smile12345 · Aug 4, 2013

Drongoski said:
When I did the question, I thought I saw the the range for theta from 0 to 180.

No worries, it's fine... All is good Drongoski

Hypem · Aug 4, 2013

Smile12345 said:
So for a more simple one like

$$Solve for 0^{\circ} \leq \Theta \leq 360^{\circ}$
$sin({\Theta}) = - Root 3 over 2$$

Would there just be two answers... 240 and 360?

240, 300 for this one

Negative, so it's in the tan+cos quadrant

180+60 = 240
360-60 = 300

Smile12345 · Aug 4, 2013

I think I'm getting the hang of it... just one more please... (After the one I believe has two solutions... )

Thanks all.

Smile12345 · Aug 4, 2013

Hypem said:
240, 300 for this one

Negative, so it's in the tan+cos quadrant

180+60 = 240
360-60 = 300

Yes, I follow you, Thanks.

(My error - taking the answer from my lined paper to here!!!)

Smile12345 · Aug 4, 2013

Smile12345 · Aug 4, 2013

Sorry I've just changed the question... It'll make a large difference. Please forgive me.

Smile12345 · Aug 4, 2013

Smile12345 said:
$$Solve for -180^{\circ} \leq \Theta \leq 180^{\circ}$
$tan^2(2{\Theta}) = 3$$

So Tan 2(Theta) = + and - root 3...

$\therefore $ over the range: $ 2\theta = \pm 60^{\circ}, \pm 120^{\circ}, \pm 240^{\circ}, \pm 300^{\circ}$

$\therefore $ = : $ \theta = \pm 30^{\circ}, \pm 60^{\circ}, \pm 120^{\circ}, \pm 150^{\circ}$

cookiez69 · Aug 4, 2013

Smile12345 said:
$$Solve for -180^{\circ} \leq \Theta \leq 180^{\circ}$
$tan^2(2{\Theta}) = 3$$

$tan(2\theta) = \pm \sqrt{3} \\ 2\theta = 60^\circ, (180^\circ-60^\circ), 240^\circ, 300^\circ, -(180^\circ-60^\circ), -60^\circ, -300^\circ, -240^\circ \\ \therefore \theta = 30, 60, 120, 150, -60, -30, -150, -120 \\ (\pm 30, \pm 60, \pm 120, \pm 150)$

I think that's correct I'm not too sure.

Trig. Help (1 Viewer)

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What a stupid name, Nat.

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What a stupid name, Nat.

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