Trig equation, is this correct?

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July 18th 2009, 10:34 AM
Tweety

Trig equation, is this correct?

Quote:

Solve the following equations in the given intervals.

$sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1$

$0\leq \theta \leq 2\pi$

$sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1$

$1 + tan^2 \theta = sec^2 \theta$

$1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1$

$tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

$tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3}$

$tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}$

so

$tan \theta = -\sqrt{3}$

or

$tan \theta = 1$

is this method correct? thanks!
July 18th 2009, 10:49 AM
Amer

Quote:

Originally Posted by Tweety

$sec^{2} \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1$

$1 + tan^2 \theta = sec^2 \theta$

$1 + tan^2 \theta - ( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 1$

$tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

$tan^2 \theta - tan \theta - \sqrt{3}tan \theta = -\sqrt{3}$

$tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}$

so

$tan \theta = -\sqrt{3}$

or

$tan \theta = 1$

is this method correct? thanks!

it is wrong
let $tan(\theta)=t$ and solve it by $t=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

then

$tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$

$\theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)$
July 18th 2009, 01:03 PM
Tweety

Quote:

Originally Posted by Amer

it is wrong
let $tan(\theta)=t$ and solve it by $t=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ in your question a=1 , b=-(1+sqrt(3)) , c=sqrt(3)

then

$tan(\theta)=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$

$\theta = \tan^{-1}\left(\frac{-b\mp \sqrt{b^2-4ac}}{2a}\right)$

Hi,

I dont understand how you got to the quadratic equation?

the correct answers are, $\frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3}$
July 18th 2009, 01:14 PM
e^(i*pi)

$tan \theta ( tan \theta-\sqrt{3} -1 ) = -\sqrt{3}<br />$

You can't factorise when it's like that, you can only factorise when one side is equal to 0

$<br /> tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0<br />$

You can use the quadratic equation from there. If it helps let $u = tan \theta$
July 18th 2009, 01:14 PM
Amer

Quote:

Originally Posted by Tweety

Hi,

I dont understand how you got to the quadratic equation?

the correct answers are, $\frac{\pi}{4} , \frac{\pi}{3} , \frac{5\pi}{4} , \frac{4\pi}{3}$

$tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

let t=tan(theta)

$t^2 -(1+\sqrt{3})t+\sqrt{3}=0$

$t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}$

$t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}$

$t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}$
July 18th 2009, 01:21 PM
Tweety

Quote:

Originally Posted by Amer

$tan^2 \theta -( 1 + \sqrt{3} )tan \theta + \sqrt{3} = 0$

let t=tan(theta)

$t^2 -(1+\sqrt{3})t+\sqrt{3}=0$

$t=\frac{1+\sqrt{3}\mp \sqrt{1+2\sqrt{3}+3-4(\sqrt{3})}}{2}$

$t=\frac{1+\sqrt{3}\mp \sqrt{1-2\sqrt{3}+3}}{2}$

$t=\frac{1+\sqrt{3}\mp \sqrt{(1-\sqrt{3})^2}}{2}$

Oh right, I get it now thanks
July 18th 2009, 02:30 PM
skeeter

$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$

$(\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$

$\tan{\theta} = \sqrt{3}$ ... $\theta = \frac{\pi}{3}$ , $\frac{4\pi}{3}<br />$

$\tan{\theta} = 1$ ... $\theta = \frac{\pi}{4}$ , $\frac{5\pi}{4}$
July 18th 2009, 02:52 PM
Tweety

Quote:

Originally Posted by skeeter

$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$

$(\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$

$\tan{\theta} = \sqrt{3}$ ... $\theta = \frac{\pi}{3}$ , $\frac{4\pi}{3}<br />$

$\tan{\theta} = 1$ ... $\theta = \frac{\pi}{4}$ , $\frac{5\pi}{4}$

how did you get this expression $(\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
from
$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$

? is it a difference of two square??
July 18th 2009, 02:58 PM
HallsofIvy

Quote:

Originally Posted by Tweety

how did you get this expression $(\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
from
$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$

? is it a difference of two square??

Suppose you had $x^2- (1+ a)x+ a$ . Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?
July 18th 2009, 03:00 PM
Amer

Quote:

Originally Posted by Tweety

how did you get this expression $(\tan{\theta} - \sqrt{3})(\tan{\theta} - 1) = 0$
from
$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$

? is it a difference of two square??

look at this

$x^2-5x+6 \Rightarrow (x-2)(x-3)$ -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

$x^2-4x+3\Rightarrow (x-1)(x-3)$ -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

like this or use

$x=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ if you do not understand it
July 18th 2009, 03:04 PM
pickslides

Difference of 2 squares here is not needed.

Skeeter has used the fact that $-\sqrt{3}\times -1 = \sqrt{3}$

and that $-\sqrt{3}+ -1 = -(\sqrt{3}+1)$ like what you would do when solving an ordinary quadratic that has 1 as the coeffecient for the $x^2$ term
July 18th 2009, 03:17 PM
Tweety

Quote:

Originally Posted by HallsofIvy

Suppose you had $x^2- (1+ a)x+ a$ . Could you see that it factors as (x-1)(x-a)? Do you see that a factors as (-1-)(a) and those add to -(1+ a), the coefficient of x?

if you had the expression $(x-1)(x-a)$ this expands out to $x^2-ax-x +a$

right?

Than you can't really factor it any further to get back to the original expresssion. which is why I can't understand how it turn outs to be (x-1)(x-a)
July 18th 2009, 03:25 PM
Tweety

Quote:

Originally Posted by Amer

look at this

Quote:

Originally Posted by Amer

$x^2-5x+6 \Rightarrow (x-2)(x-3)$ -2,-3 the multiply of them is 6 and the sum is -5 . 6 is the constant term and -5 is the factor of x

$x^2-4x+3\Rightarrow (x-1)(x-3)$ -1,-3 multiply of them is 3 and the sum is -4 .3 is the constant term and -5 is the factor of x

like this or use

$x=\frac{-b\mp \sqrt{b^2-4ac}}{2a}$ if you do not understand it

I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$ As I would with $x^2-5x+6$ ?

thanks.
July 18th 2009, 03:35 PM
skeeter

Quote:

Originally Posted by Tweety

I understand your examples but how does this relate to the trig expression, are you saying that I can use the same method to factorize
$\tan^2{\theta} -( 1 + \sqrt{3} )\tan{\theta} + \sqrt{3} = 0$ As I would with $x^2-5x+6$ ?

thanks.

yes ...

$x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6$

the only difference here is that $3$ and $2$ can be combined, wheras the $1$ and $\sqrt{3}$ cannot.
July 18th 2009, 03:39 PM
Tweety

Quote:

Originally Posted by skeeter

yes ...

$x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x^2 - (3+2)x + 6$

the only difference here is that $3$ and $2$ can be combined, wheras the $1$ and $\sqrt{3}$ cannot.

oh right finally I get it, that makes so much sense now!!

Thanks!

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