# Extremely Urgent Trig Help Needed

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• Jan 2nd 2008, 07:00 PM
Gerbilkit
Extremely Urgent Trig Help Needed
I have an assignment to turn in tomorrow that will be a large part of my grade. I have had virtually no time to work over break and many other things to work on. Yada yada yada everyone rationalizes why they didn't get stuff done so it doesn't really matter why. I need to know how to do these problems by tomorrow morning. If someone was kind enough to type out the explanations I could finish it during lunch easily, I used to know how but forgot over a couple weeks, guess that goes to show my learning wasn't very deep.

So here are the problems I will need to know by tomorrow. I appreciate anyone who can help me out right away.

1. $\displaystyle \sin 2 \theta \ = 1/2$

2. $\displaystyle 3 \sqrt{2} \sin \theta + 2 = -1$

3. $\displaystyle 2 \sin^2 \ theta\ - 3 \cos \theta +1 = 0$

4. $\displaystyle \cos 2\ theta = cos \ theta \$

Any and all help right away is very much appreciated. As I said this is really important and urgent.
• Jan 2nd 2008, 07:13 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
I have an assignment to turn in tomorrow that will be a large part of my grade. I have had virtually no time to work over break and many other things to work on. Yada yada yada everyone rationalizes why they didn't get stuff done so it doesn't really matter why. I need to know how to do these problems by tomorrow morning. If someone was kind enough to type out the explanations I could finish it during lunch easily, I used to know how but forgot over a couple weeks, guess that goes to show my learning wasn't very deep.

So here are the problems I will need to know by tomorrow. I appreciate anyone who can help me out right away.

1. Sin(2 theta) = 1/2

what range for $\displaystyle \theta$ do you want to solve for?

you should know that (in general) if $\displaystyle \sin x = \frac 12$ then $\displaystyle x = \frac {\pi}6 + 2k \pi$ or $\displaystyle x = \frac {5 \pi}6 + 2n \pi$ for $\displaystyle k,n \in \mathbb{Z}$

so, $\displaystyle \sin 2 \theta = \frac 12 \implies 2 \theta = \frac {\pi}6 + 2k \pi$ or $\displaystyle 2 \theta = \frac {5 \pi}6 + 2n \pi$ for $\displaystyle k,n \in \mathbb{Z}$

now find $\displaystyle \theta$
• Jan 2nd 2008, 07:16 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
2. 3 square root of 2 sin(theta) + 2 = -1

kinda confused here... did you mean $\displaystyle 3 \sqrt{2 \sin \theta} + 2 = -1$ ?
• Jan 2nd 2008, 07:17 PM
Gerbilkit
Sorry forgot to add these should all be solved on the interval of 0 <= theta <= 2pi.

Could you elaborate slightly? My memory of the periodic functions is a bit hazy at the moment. Although I am looking at some more examples currently. Also my brain is a bit fried at the moment.

And yes that was exactly what I meant, sorry I'm not great with the math tags having just started.
• Jan 2nd 2008, 07:20 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
3. 2sin squared theta - 3 cos(theta) +1 = 0

recall that $\displaystyle \sin^2 \theta = 1 - \cos^2 \theta$, so

$\displaystyle 2 \sin^2 \theta - 3 \cos \theta + 1 = 0$

$\displaystyle \Rightarrow 2 - 2 \cos^2 \theta - 3 \cos \theta + 1 = 0$

$\displaystyle \Rightarrow 2 \cos^2 \theta + 3 \cos \theta - 3 = 0$

now this is quadratic in $\displaystyle \cos \theta$, if we say replace $\displaystyle \cos \theta$ with $\displaystyle x$, we get:

$\displaystyle 2x^2 + 3x - 3 = 0$

we can solve for $\displaystyle x$ here, once we get $\displaystyle x$, just replace it with $\displaystyle \cos \theta$ and solve for $\displaystyle \theta$
• Jan 2nd 2008, 07:23 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
4. cos (2theta) = cos (theta)

here, use the fact that $\displaystyle \cos 2 \theta = 2 \cos^2 \theta - 1$. plug that in for $\displaystyle \cos 2 \theta$ and then you would solve it exactly like number 3
• Jan 2nd 2008, 07:30 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
Sorry forgot to add these should all be solved on the interval of 0 <= theta <= 2pi.

in that case, $\displaystyle 2 \theta = \frac {\pi}6$ or $\displaystyle \frac {5 \pi}6$

Quote:

Could you elaborate slightly? My memory of the periodic functions is a bit hazy at the moment. Although I am looking at some more examples currently. Also my brain is a bit fried at the moment.
hmm, that will be a pain. how is your knowledge of reference angles or of the graphs of sine and cosine?

Quote:

And yes that was exactly what I meant, sorry I'm not great with the math tags having just started.
ok then,

$\displaystyle 3 \sqrt{2 \sin \theta} + 2 = -1$ .............we are going to solve for sine. subtract 2 from both sides

$\displaystyle \Rightarrow 3 \sqrt{2 \sin \theta } = -3$ ...........divide by 3 on both sides

$\displaystyle \Rightarrow \sqrt{2 \sin \theta } = -1$ .............square both sides

$\displaystyle \Rightarrow 2 \sin \theta = 1$ .............divide by 2 on both sides

$\displaystyle \Rightarrow \sin \theta = \frac 12$

which is the same as question 1

(by the way, be sure to check your solutions in the original equation. sometimes when you square things and such, you introduce erroneous solutions).

of course we can tell that this equation has no solution from the get-go. so doing all of this, while you would want to show your work, is a waste of time. why is that?
• Jan 2nd 2008, 07:42 PM
Gerbilkit
Quote:

Originally Posted by Jhevon
here, use the fact that $\displaystyle \cos 2 \theta = 2 \cos^2 \theta - 1$. plug that in for $\displaystyle \cos 2 \theta$ and then you would solve it exactly like number 3

I solved that out and factored it, and I got cos = -1/2 and cos = 1. Right now I'm thinking cos = 1 or 2 pi is the answer. Is that right?

I'm sorry, I messed up when I wrote that to you. The sin value is not actually under the square root, only the two. So it doesn't quite work out the way you did it.
• Jan 2nd 2008, 07:47 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
I solved that out and factored it, and I got cos = -1/2 and cos = 1. Right now I'm thinking cos = 1 or 2 pi is the answer. Is that right?

yes, $\displaystyle \cos \theta = - \frac 12$ or $\displaystyle \cos \theta = 1$. but after that, you are wrong. you are supposed to find $\displaystyle \theta$, for what values of $\displaystyle \theta$ do the above happen?

Quote:

Originally Posted by Gerbilkit
I'm sorry, I messed up when I wrote that to you. The sin value is not actually under the square root, only the two. So it doesn't quite work out the way you did it.

ok, then you tell me where to go with it this time. the plan of attack is the same. we are going to isolate the $\displaystyle \sin \theta$. so what's the equation? what does $\displaystyle \sin \theta =$ ?
• Jan 2nd 2008, 07:50 PM
Gerbilkit
The true equation is.

$\displaystyle 3 \sqrt{2} \sin \theta + 2 = -1$

All the others should be correct.
• Jan 2nd 2008, 07:55 PM
Jhevon
Quote:

Originally Posted by Gerbilkit
The true equation is.

$\displaystyle 3 \sqrt{2} \sin \theta + 2 = -1$

All the others should be correct.

yes. now solve for $\displaystyle \sin \theta$
• Jan 3rd 2008, 07:43 AM
Gerbilkit
I worked out everything now for the most part except for these two. The sin of 2 theta = 1/2 one. Please just work this out someone, I really tried to remember and work out periodic properties but I'm failing atm.

Also I tried that one that worked out to a quadratic and didn't factor.

$\displaystyle 2x^2 + 3x - 3 = 0$

I put it through the quadratic equation and came up with.

$\displaystyle - 3 +- \sqrt{33}/ \ 4$

Did I do it wrong? I can't come up with an exact value of theta for this junk.
• Jan 3rd 2008, 07:49 AM
colby2152
Quote:

Originally Posted by Gerbilkit
I worked out everything now for the most part except for these two. The sin of 2 theta = 1/2 one. Please just work this out someone, I really tried to remember and work out periodic properties but I'm failing atm.

$\displaystyle sin(2\theta)=\frac{1}{2}$

Let $\displaystyle y = 2\theta$

$\displaystyle sin(y)=\frac{1}{2}$

$\displaystyle y = \frac{\pi}{6}$

$\displaystyle 2\theta = \frac{\pi}{6}$

$\displaystyle \theta = \frac{\pi}{12}$
• Jan 3rd 2008, 07:54 AM
colby2152
Quote:

Originally Posted by Gerbilkit
Also I tried that one that worked out to a quadratic and didn't factor.

$\displaystyle 2x^2 + 3x - 3 = 0$

I put it through the quadratic equation and came up with.

$\displaystyle - 3 +- \sqrt{33}/ \ 4$

Did I do it wrong? I can't come up with an exact value of theta for this junk.

Your solution to the quadratic looks fine. You could not have factored it, and were forced to use the quadratic formula.
• Jan 3rd 2008, 12:18 PM
Jhevon
Quote:

Originally Posted by colby2152
$\displaystyle sin(2\theta)=\frac{1}{2}$

Let $\displaystyle y = 2\theta$

$\displaystyle sin(y)=\frac{1}{2}$

$\displaystyle y = \frac{\pi}{6}$

$\displaystyle 2\theta = \frac{\pi}{6}$

$\displaystyle \theta = \frac{\pi}{12}$

we want $\displaystyle 0 \le \theta \le 2 \pi$ so $\displaystyle \theta = \frac {5 \pi}{12}$ is the other solution
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