# 2sinx-cotxcosx=1

• February 24th 2011, 12:24 PM
jonnygill
2sinx-cotxcosx=1
the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

after switching terms around and doing some operations i got...

$4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si n^6\Theta}$

This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

Is there another, easier approach?

thanks!
• February 24th 2011, 12:40 PM
topsquark
Quote:

Originally Posted by jonnygill
the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

-Dan
• February 24th 2011, 01:06 PM
jonnygill
Quote:

Originally Posted by topsquark
$\displaystyle 2 sin( \theta ) - cot( \theta) cos( \theta ) = 0$

$\displaystyle 2 sin( \theta ) - \frac{cos^2( \theta)}{sin(\theta )} = 0$

$\displaystyle 2 sin^2( \theta ) - cos^2( \theta) = 0$

Now use $\displaystyle sin^2( \theta ) = 1 - cos^2( \theta )$

-Dan

could you show how you went from line 2 to line 3?

i tried getting a common denominator but i ended up with $-\dfrac{1}{sin^3\Theta}+2sin\Theta$

also, where did the 1 go in the original equation? why did you just remove it?

edit: i see how you went from line 2 to line 3. but i do not understand why you set the equation equal to zero and not one. was this an oversight?
• February 24th 2011, 01:19 PM
jonnygill
i kept the one and got $3sin^2\Theta-sin\Theta-1=0$

I can now use the quad. formula to solve for sin(Theta). I suspect you removed the 1 on purpose. perhaps it makes it easier to solve. but how would you re-incorporate that 1 back into the equation when you are finished?

edit: i solved the equation using the quad. formula and got two answers that work with the original equation. Thanks!
• February 24th 2011, 01:19 PM
Quote:

Originally Posted by jonnygill
the problem is...

$2sin\Theta-cot\Theta cos\Theta=1$

after switching terms around and doing some operations i got...

$4sin^2\Theta=1+\dfrac{2}{sin^3\Theta}+\dfrac{1}{si n^6\Theta}$

This feels like the wrong direction since i'm using terms with sin^6 in the denominator.

When i first looked at it, it seemed like getting all the terms to be sin (or cos too i guess) would be possible. Well i got all the terms to be in sin, but these terms have large exponents in the denominator, etc.

Is there another, easier approach?

thanks!

Simply use

$\displaystyle\frac{cos\Theta}{sin\Theta}=cot\Theta$

which gives

$\displaystyle\ 2sin\Theta-\frac{cos\Theta}{sin\Theta}\;cos\Theta=1\Rightarro w\left[\frac{sin\Theta}{sin\Theta}\right]2sin\Theta-\frac{cos^2\Theta}{sin\Theta}=1$

which gives a common denominator

$\displaystyle\frac{2sin^2\Theta-cos^2\Theta}{sin\Theta}=1$

$2sin^2\Theta-\left(1-sin^2\Theta\right)=sin\Theta$