# Thread: Help with trig algebra. I can't figure this out

1. ## Help with trig algebra. I can't figure this out

Ok, so I am looking at the min and max for the angle x:

-12cos(x) + (2)12sin(x) = 5

and

-12cos(x) + (2)12sin(x) = -5

which I reduced to:

(12)(2sin(x) - cos(x)) = ±5

2sin(x) - cos(x) = ±5/12

I'm not sure how to solve for the angle (x) from here.

2. Originally Posted by mathdude09
Ok, so I am looking at the min and max for the angle x:

-12cos(x) + (2)12sin(x) = 5

and

-12cos(x) + (2)12sin(x) = -5

which I reduced to:

(12)(2sin(x) - cos(x)) = ±5

2sin(x) - cos(x) = ±5/12

I'm not sure how to solve for the angle (x) from here.
I'm not sure about what is going on here, but:

$12(2\sin{x}-\cos{x})=5$

$2\sqrt{1-\cos^2{x}}-\cos{x}=\frac{5}{12}$

$2\sqrt{1-cos^2{x}}=\frac{5}{12}+\cos{x}$

Squaring both sides

$4-4\cos^2{x}=\frac{25}{144}+\frac{5}{6}\cos{x}+\cos^ 2{x}$

Now you have a quadratic in $cos{x}$.

3. Working with what you gave me I was able to successfully get both angles using the quadratic formula, thanks.

Just to understand exactly how you manipulated the sin function:

sin(x) = sqrt(1-cos(x)^2) ???

Is there a listing of these conversions somewhere?

4. Originally Posted by mathdude09
Working with what you gave me I was able to successfully get both angles using the quadratic formula, thanks.

Just to understand exactly how you manipulated the sin function:

sin(x) = sqrt(1-cos(x)^2) ???

Is there a listing of these conversions somewhere?
This was derived from the pythagorean identity IE

$\sin^2{x}+\cos^2{x}=1$

Subtracting...

5. Oh yeah. Pythagoras, unit circle...it's all coming back now. Thanks