Another Trig Equation


by cscott
Tags: equation, trig
cscott
cscott is offline
#1
Oct6-05, 03:24 PM
P: 786
[tex]
\begin{align*}
\sin^2 \theta + 2 \sin \theta \cos \theta - 3 \cos^2 \theta = 0 \\
(\sin \theta + 3 \cos \theta)(\sin \theta - \cos \theta) = 0 \\
\end{align*}
[/tex]

so,

[tex]
\begin{align*}
\sin \theta = \cos \theta\\
\theta = \frac{\pi}{4} + \pi k, k \epsilon \mathbb{I}
\end{align*}
[/tex]

or

[tex]
\begin{align*}
\sin \theta = -3 \cos \theta\\
\tan \theta = -3\\
\theta = 5.03 + 2\pi k, k \epsilon \mathbb{I}; 1.89 + 2\pi k, k \epsilon \mathbb{I}
\end{align*}
[/tex]

How come this is incorrect?

...argh I can't align it properly
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hotvette
hotvette is offline
#2
Oct6-05, 03:46 PM
HW Helper
P: 930
a x b = 0 doesn't mean both a and b are zero, although they could be. It simply means at least one of them is zero for the product to be zero.
TD
TD is offline
#3
Oct6-05, 04:22 PM
HW Helper
P: 1,024
Quote Quote by cscott
[tex]\begin{align*}
\sin \theta = \cos \theta\\
\theta = \frac{\pi}{4} + 2\pi k, k \epsilon \mathbb{I}
\end{align*}
[/tex]
Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.

cscott
cscott is offline
#4
Oct6-05, 04:34 PM
P: 786

Another Trig Equation


Quote Quote by TD
Here, you can get all solution by taking pi/4 + k*pi instead of 2k*pi.
Oops, I had that written down but I made the mistake when posting!

Does this mean you say pi/4 is a correct answer? When I plug it back into the original equation I don't get 0
TD
TD is offline
#5
Oct6-05, 04:38 PM
HW Helper
P: 1,024
How's that?

[tex]
\sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0
[/tex]
cscott
cscott is offline
#6
Oct6-05, 04:48 PM
P: 786
Quote Quote by TD
How's that?

[tex]
\sin ^2 \theta + 2\sin \theta \cos \theta - 3\cos ^2 \theta \mathop \to \limits^{\theta = \pi /4} \left( {\sin \frac{\pi }{4}} \right)^2 + 2\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 3\left( {\cos \frac{\pi }{4}} \right)^2 = \frac{1}{2} + 2 \cdot \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} - 3 \cdot \frac{1}{2} = 0
[/tex]
OK, I'm just bad at typing things into my calculator then! That's what I get for using it in the first place

By the way, can anyone tell me how to align things properly with tex? For some reason using "\\" won't skip lines for me.

Thanks.
TD
TD is offline
#7
Oct6-05, 04:59 PM
HW Helper
P: 1,024
It will if you use an array:

[tex]\begin{array}{l}
x^2 - 4 = 0 \\
x = 2 \vee x = - 2 \\
\end{array}[/tex]

\begin{array}{l}
 x^2  - 4 = 0 \\ 
 x = 2 \vee x =  - 2 \\ 
 \end{array}
cscott
cscott is offline
#8
Oct6-05, 05:39 PM
P: 786
Doh! Thanks again!
TD
TD is offline
#9
Oct6-05, 05:39 PM
HW Helper
P: 1,024
You're welcome


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