1. ## HELPPPP! Trigonometry!

2(sin^-1x)cubed+5(sin^-1x)squared-1=0

As well as find the area and perimeter of the triangle.

cos(a) equals approximately -.125, b=8, c=8

Helpppp!

-Thanks

2. Originally Posted by weehoo16
2(sin^-1x)cubed+5(sin^-1x)squared-1=0
Let $\displaystyle y = sin^{-1}(x)$

Then we need to solve
$\displaystyle 2y^3 + 5y - 1 = 0$

The rational root theorem says that if there is a rational root for this polynomial equation it is of the form $\displaystyle y = \pm 1, \pm \frac{1}{2}$.

Trial and error shows that y = -1/2 is indeed a solution. This means we can draw out a factor of the form 2y + 1:
$\displaystyle 2y^3 + 5y - 1 = (2y + 1)(y^2 + 2y - 1)$
by any method of polynomial division you choose.

So finishing the solution I get that
$\displaystyle y = -\frac{1}{2}, 1 \pm \sqrt{2}$

So we need to solve for the three solutions:
$\displaystyle x = sin^{-1} \left ( -\frac{1}{2} \right )$
and
$\displaystyle x = sin^{-1}(1 + \sqrt{2} )$
and
$\displaystyle x = sin^{-1}(1 - \sqrt{2} )$

The last two are impossible, as $\displaystyle |1 \pm \sqrt{2}| > 1$. So we only have
$\displaystyle x = sin^{-1} \left ( -\frac{1}{2} \right ) \implies x = -\frac{\pi}{6}~rad$

-Dan

3. Originally Posted by weehoo16
2(sin^-1x)cubed+5(sin^-1x)squared-1=0

As well as find the area and perimeter of the triangle.

cos(a) equals approximately -.125, b=8, c=8

Helpppp!

-Thanks
To get the third side of the triangle, use the Law of Cosines. Then you can get the perimeter easily.

To get the area you can use Heron's formula.

-Dan

4. Originally Posted by topsquark
To get the third side of the triangle, use the Law of Cosines. Then you can get the perimeter easily.

To get the area you can use Heron's formula.

-Dan
So the answer to the question would be 0. Yeah?

5. Originally Posted by weehoo16
So the answer to the question would be 0. Yeah?
What would be 0? I'm getting that the length of the third side is 12:
$\displaystyle a^2 = b^2 + c^2 - 2bc \cdot cos(A)$

-Dan