Author Topic: Holiday homeworkk (Read 10651 times) Tweet Share

mandy · « **Reply #60 on:** December 31, 2009, 03:14:49 pm »

Solve the following equation for $x --> [0, 2\pi]$
sinx + cosx = 1

How do I do this question?

Dark Horse · « **Reply #61 on:** December 31, 2009, 06:12:33 pm »

square both sides

you get (sinx)^2 + (cosx)^2 + 2sinxcosx = 1
2sinxcosx=0
sin2x = 0
x = 0, pie/2

Sorry the answer is shotty, I dont know how to use latex

and can someone please check, I cant guarantee this is right

mandy · « **Reply #62 on:** December 31, 2009, 08:33:46 pm »

Dark Horse, thanks for your working out

Another answer was $2\pi$ .

mandy · « **Reply #63 on:** January 01, 2010, 01:36:28 pm »

Can someone check my answer for this?

Solve the following eqn over $[0,2\pi]$ .
$sec x = 2$
Does x = $\frac{\pi}{3} , \frac{5\pi}{3}$ ?

Also, I've forgotten how to do this:

Find the exact value of: $tan\frac{7\pi}{12}$ .
Can you give me a hint or something :S

Aden · « **Reply #64 on:** January 01, 2010, 01:57:08 pm »

Yes, you're first question works. As for the second one, use this compound angle formula:

$tan(x+y) = \frac{tan(x) + tan(y)}{1 - tan(x)*tan(y)}$

Where 'x' can be $\frac{\pi}{3}$ and 'y' can be $\frac{\pi}{4}$ .

mandy · « **Reply #65 on:** January 01, 2010, 02:48:09 pm »

Thank you Aden, I got the answer

mandy · « **Reply #66 on:** January 01, 2010, 02:54:23 pm »

Oh no, what do I do for this one:
Find the exact value of $cosec \frac{\pi}{12}$ ?

GerrySly · « **Reply #67 on:** January 01, 2010, 03:03:28 pm »

$\begin{align*} \csc{(\frac{\pi}{12})}&=\frac{1}{\sin{(\frac{\pi}{12})}}\\ &=\frac{1}{\sin{(\frac{\pi}{3}-\frac{\pi}{4})}}\\ \end{align*}$

Apply the compound angle thinga-mi-bob and boom you've got your answer

mandy · « **Reply #68 on:** January 01, 2010, 03:22:29 pm »

This is my working out:
$sin(x - y) = sinx cosy - cosx siny$
where $x = \frac{\pi}{3}$ and $y = \frac{\pi}{4}$

$sin(\frac{\pi}{12})$ = $sin(\frac{\pi}{3}) cos(\frac{\pi}{4}) - cos(\frac{\pi}{3}) sin(\frac{\pi}{4})$

= $(\frac{\sqrt3}{2} X \frac{\sqrt2}{2})$ - $(\frac{1}{2} X \frac{\sqrt2}{2})$

= $\frac{\sqrt6}{4} - \frac{\sqrt2}{4}$

= $\frac{\sqrt6 - \sqrt2}{4}$

And because it's $cosec$ , I reciprocated the answer.
So, I got $\frac{4}{\sqrt6 - \sqrt2}$ .

Is that correct?

GerrySly · « **Reply #69 on:** January 01, 2010, 03:41:01 pm »

Quote from: mandy on January 01, 2010, 03:22:29 pm

So, I got $\frac{4}{\sqrt6 - \sqrt2}$ .

Is that correct?

Yes, but you can go further than that.

$\begin{align*} \frac{4}{\sqrt{6}-\sqrt{2}}&=\frac{4}{\sqrt{6}-\sqrt{2}}\cdot\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}\\ &=\frac{4(\sqrt{6}+\sqrt{2})}{4}\\ &=\sqrt{6}+\sqrt{2} \end{align*}$

mandy · « **Reply #70 on:** January 01, 2010, 03:54:01 pm »

Thank you GerrySly!

I have another problem.

Simplify each of the following:

$cos(\frac{3\pi}{2} + x)$

I saw that rule that says: $cos(\frac{3\pi}{2} + \theta) = sin\theta$
But does this help me in any way?

mandy · « **Reply #71 on:** January 01, 2010, 04:44:13 pm »

Anyone with a clue ?

GerrySly · « **Reply #72 on:** January 01, 2010, 04:53:08 pm »

Quote from: mandy on January 01, 2010, 04:44:13 pm

Anyone with a clue ?

You answered it yourself...

$\cos{(\frac{3\pi}{2}+x)}&=\sin{(x)}$

mandy · « **Reply #73 on:** January 01, 2010, 04:55:24 pm »

OH, LOL. IS THAT IT ==
THANK YOU

mandy · « **Reply #74 on:** January 01, 2010, 05:22:23 pm »

I'm having a lot of trouble understanding how to do these kinds of problems. For example, I look at this next question, and it looks like a rule, but it's like ... backwards.

Simplify the following:
a. $cot (\frac{\pi}{2} - y)$
For this one, the rule says: $tan(\frac{\pi}{2} - \theta) = cot x$
What am I meant to do?

b. $cosec(2\pi - x)$
And for this one, there isn't anything about $cosec$ in my rules at all :S

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