# Finding a trig function given the value of another

• Apr 23rd 2009, 09:45 PM
Josh_629
Finding a trig function given the value of another
I have been trying to solve this equation for a long time, can someone please help me? Find cos 2x if you are in quadrant 4 and cot x is -7/24. I'm not sure which equation to use since I am given cotangent rather then sine or cos.
• Apr 23rd 2009, 09:54 PM
Jhevon
Quote:

Originally Posted by Josh_629
I have been trying to solve this equation for a long time, can someone please help me? Find cos 2x if you are in quadrant 4 and cot x is -7/24. I'm not sure which equation to use since I am given cotangent rather then sine or cos.

recall, cot(x) = 1/tan(x)

also note that $\displaystyle \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$. you only need to use one of these alternate forms.

now, draw a right triangle with an acute angle x, since cot(x) = adjacent/hypotenuse, you know the sides opposite and adjacent to this angle. use Pythagoras' theorem to find the other side. then you can find sin(x) and cos(x) and hence cos(2x). note you are in the fourth quadrant, so sine is negative, which cosine is positive. this doesn't really matter though, since we are squaring...
• Apr 23rd 2009, 10:20 PM
Josh_629
That's as far as I got, but I'm not really sure how to solve it from there. Which one of the three equations should I use? That part I am not too sure of.
• Apr 23rd 2009, 10:26 PM
Prove It
Quote:

Originally Posted by Josh_629
I have been trying to solve this equation for a long time, can someone please help me? Find cos 2x if you are in quadrant 4 and cot x is -7/24. I'm not sure which equation to use since I am given cotangent rather then sine or cos.

In quadrant 4, $\displaystyle \tan{x} = -\frac{24}{7}$

Also note that $\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$

So $\displaystyle \frac{\sin{x}}{\cos{x}} = -\frac{24}{7}$

$\displaystyle \frac{\sin^2{x}}{\cos^2{x}} = \frac{24^2}{7^2}$

$\displaystyle \frac{1 - \cos^2{x}}{\cos^2{x}} = \frac{24^2}{7^2}$

$\displaystyle 1 - \cos^2{x} = \left(\frac{24^2}{7^2}\right)\cos^2{x}$

Now collect the like terms and use the fact that $\displaystyle \cos^2{x} = \frac{1}{2} + \frac{1}{2}\cos{(2x)}$ to solve for $\displaystyle \cos{(2x)}$.

$\displaystyle 1 - \cos^2{x} = \left(\frac{24^2}{7^2}\right)\cos^2{x}$

$\displaystyle 1 = \left(\frac{24^2}{7^2}\right)\cos^2{x} + \cos^2{x}$

$\displaystyle 1 = \left(\frac{625}{49}\right)\cos^2{x}$

$\displaystyle \frac{49}{625} = \cos^2{x}$

$\displaystyle \frac{49}{625} = \frac{1}{2}\left[1 + \cos{(2x)}\right]$

$\displaystyle \frac{98}{625} = 1 + \cos{(2x)}$

$\displaystyle -\frac{527}{625} = \cos{(2x)}$
• Apr 23rd 2009, 10:30 PM
Josh_629
Is the solution 576/625? I think I did the math correctly, can anyone verify this?