1. ## trigonometry simplify

$\displaystyle 750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2$

Find the value of x.

2. Originally Posted by iphysics
$\displaystyle 750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2$

Find the value of x.

Well, sin(1) is just a constant, so
$\displaystyle \frac{750}{\pi} \cdot 48^2 = \frac{sin^2(1 + x) - sin^2(1)}{2}$

$\displaystyle 2 \cdot 48^2 \cdot \frac{750}{\pi} = sin^2(1 + x) - sin^2(1)$

$\displaystyle sin^2(1 + x) = 2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)$

$\displaystyle sin(1 + x) = \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)}$

$\displaystyle x + 1 = sin^{-1} \left ( \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)} \right )$

$\displaystyle x = -1 + sin^{-1} \left ( \pm \sqrt{2 \cdot 48^2 \cdot \frac{750}{\pi} + sin^2(1)} \right )$

Unfortunately $\displaystyle 2 \cdot 48^2 \cdot 750/ \pi + sin^2(1)$ is rather larger than 1 so the value of x is imaginary, which is a result I'm assuming you don't want. So there is no real solution to this problem.

-Dan

3. Originally Posted by iphysics
$\displaystyle 750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2$

Find the value of x.

$\displaystyle [\sin(1+x)]^2=\frac{1500}{48^2 \pi}+ \sin(1)$

so:

$\displaystyle \sin(1+x)=\sqrt{\frac{1500}{48^2 \pi}+ \sin(1)}$

Then:

$\displaystyle x=\arcsin \left( \sqrt{\frac{1500}{48^2 \pi}+ \sin(1)}\right)-1$

RonL

4. Originally Posted by iphysics
$\displaystyle 750/ \pi\*48^2 = ((sin (1+x))^2 - (sin 1)^2)/2$
Originally Posted by CaptainBlack
$\displaystyle [\sin(1+x)]^2=\frac{1500}{48^2 \pi}+ \sin(1)$
Well that makes more sense.