Alan

1 m³ = 1000 L so 1 m³ = 1 kL

Here's the graph:

Can you answer the question now?

y=A*sin(B*x-C)+D

Here, B is 2pi/L where L is the (presumably spatial) period.  As L gets bigger the sine wave "stretches out"; as it gets smaller, the wave oscillates more rapidly.

-C is the phase shift.  When C is zero the sin wave starts at zero (and y starts at D) when x is zero.

A is the maximum amplitude.  As A gets bigger the peak to trough variation gets bigger.

D is a fixed offset.  It's the average value about which the sin wave oscillates.

$$Do you mean $$A*sin(\frac{2\pi t}{T}+\phi)$$\\
where A is maximum amplitude, t is time, T is period and $\phi$ is phase angle? The range is then $\pm A$$$

I think you might be trying to solve the following equations (from a previous problem):

17 = v1*cos(45) + v2*cos(30)  ...(1)

0 = -v1*sin(45) + v2*sin(30)     ...(2)

Add v1*sin(45) to both sides of (2) and then divide both sides by sin(45)

v1 = v2*sin(30)/sin(45)             ...(3)

Replace v1 in (1) by (3)

17 =  v2*sin(30)/sin(45)*cos(45) + v2*cos(30)

Factor the right-hand side

17 =  v2*[sin(30)/sin(45)*cos(45) + cos(30)]

Divide both sides by [sin(30)/sin(45)*cos(45) + cos(30)] to get v2

v2 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]   ...(4)

Replace v2 in (3) by (4) to get v1

v1 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]*sin(30)/sin(45)   

$${\mathtt{v1}} = {\frac{{\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}} \Rightarrow {\mathtt{v1}} = {\mathtt{8.799\: \!847\: \!533\: \!482\: \!900\: \!7}}$$

$${\mathtt{v2}} = {\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}} \Rightarrow {\mathtt{v2}} = {\mathtt{12.444\: \!863\: \!728\: \!674\: \!910\: \!2}}$$

The circumference is 94.2 inches, and circumference is pi*diameter, so diameter = 94.2/pi inches.

Using 3.14 as an approximation for pi we get:

$${\mathtt{diameter}} = {\frac{{\mathtt{94.2}}}{{\mathtt{3.14}}}} \Rightarrow {\mathtt{diameter}} = {\mathtt{30}}$$

Diameter is 30 inches.  So would need a length of at least 4*30 = 120 inches of felt for four wheels. 

The area of the circular wheel is pi*diameter²/4 square inches.

$${\mathtt{area}} = {\frac{{\mathtt{3.14}}{\mathtt{\,\times\,}}{{\mathtt{30}}}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{area}} = {\mathtt{706.5}}$$

Area is 706.5 square inches.  This much felt is needed to cover one wheel.

2sin²x - 1 = 0 means sinx = ±1/√2  Picture a right-angled triangle with sides 1, 1 and √2. The first-quadrant angle whose sin is 1/√2 is therefore 45°.  The second quadrant value is 135°.  The third and fourth quadrants have negative values, so here x is 225° and 315°

3tan²x - 1 = 0 means tanx = ±1/√3  Picture a right-angled triangle with sides 1, √3 and 2.  This triangle has angles 30° and 60°, with tan 30° = 1/√3. In the third quadrant tan is also positive, so tan 220° is also 1/√3.

In the second and fourth quadrants tan is negative so tan 150° and tan 330° are equal to -1/√3 .

It is the wavelength of light that is measured in nm not luminosity!

nm stands for nanometres.  A nanometre is 10^-9 metres.

Your solution should have 600*10^-9 in it, not 600*10⁺⁹.  The resulting thickness is then given in metres (m).  If you leave out the 10^-9  the resulting thickness would be in nanometres (nm).