# Math Help - [SOLVED] Trig. and Sin Wave Question

1. ## [SOLVED] Trig. and Sin Wave Question

If anyone can help me with these two questions I would appreciate it. Thank you!

Trig. question:

"The expression $[2cos(theta)/sin2(theta)]$ is equivalent to...?
Choices:
(1) [csc theta]
(2) [sec theta]
(3) [cot theta]
(4) [sin theta]

Work done so far:
$[Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta]$
...where to go from here?

Sin Wave Question:
"A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $[4(pie symbol)]$. Write the equation that represents this wave."

Work done so far:
Amp.= 3
Freq.= 2
Range=...5?

Equation: 3 sin 2x + ...?

Side Note: I'm new to using "LaTex"...

2. Originally Posted by MrOats
If anyone can help me with these two questions I would appreciate it. Thank you!

Trig. question:
"The expression $[2cos(theta)/sin2(theta)]$ is equivalent to...?
Choices:
(1) [csc theta]
(2) [sec theta]
(3) [cot theta]
(4) [sin theta]

Work done so far:
$[Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta]$
...where to go from here?

Sin Wave Question:
"A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $[4(pie symbol)]$. Write the equation that represents this wave."

Work done so far:
Amp.= 3
Freq.= 2
Range=...5?

Equation: 3 sin 2x + ...?

Side Note: I'm new to using "LaTex"...
For the first one

$\sin(2\theta)=2\sin(\theta)\cos(\theta)$

$\therefore\frac{2\cos(\theta)}{\sin(2\theta)}=\fra c{2\cos(\theta)}{2\sin(\theta)\cos(\theta)}$

I think you can take it from there

3. Originally Posted by MrOats
If anyone can help me with these two questions I would appreciate it. Thank you!

Trig. question:
"The expression $[2cos(theta)/sin2(theta)]$ is equivalent to...?
Choices:
(1) [csc theta]
(2) [sec theta]
(3) [cot theta]
(4) [sin theta]

Work done so far:
$[Cos2 theta/Sin2 theta] = [1-sin^2 theta/1-cos^2 theta]$
...where to go from here?

Sin Wave Question:
"A student attaches one end of a rope to a wall at a fixed point 3 feet above the ground, as shown in the accompanying diagram, and moves the other end of the rope up and down, producing a wave described by the equation $y = a sin bx + c$. The range of the rope’s height above the ground is between 1 and 5 feet. The period of the wave is $[4(pie symbol)]$. Write the equation that represents this wave."

Work done so far:
Amp.= 3
Freq.= 2
Range=...5?

Equation: 3 sin 2x + ...?

Side Note: I'm new to using "LaTex"...
It moves from 1 to 5 for the second one...so amplitude is 4

Since in the equation $a\sin(bx+c)$

The period is $\frac{2\pi}{|b|}$

So you have $\frac{2\pi}{|b|}=4\pi$

Solve for b and consider that the lowest point is one and the highest point is 5 to figure out the c value which just shifts the sine wave vertically

4. ## Please check/help me further.

Please check, someone, to make sure I did these right? Thank you!

For this question:
I got ${cos(theta)/sin(theta)}$, which is $cot(theta)$.

For the second question:
I got Amplitude = 4, Period = [2(pie)/b], Frequency = 2/4

And for the equation: y= 4 Sin (2/4(x)+4)

5. Originally Posted by MrOats
Please check, someone, to make sure I did these right? Thank you!

For this question:
I got ${cos(theta)/sin(theta)}$, which is $cot(theta)$.

No you are wrong...

$\frac{2\cos \theta}{\sin 2\theta} = \frac{2\cos \theta}{2\sin \theta\cos \theta} = \frac1{\sin\theta} = csc\theta$

Originally Posted by MrOats
For the second question:
I got Amplitude = 4, Period = [2(pie)/b], Frequency = 2/4

And for the equation: y= 4 Sin (2/4(x)+4)

I think you are misinterpreting the question ....

$y = (a\sin bx) + c$
Is the distance x from the wall or the person?

Anyway I will assume from the wall,
Note that when x = 0, then y = 3. So c = 3.

Since the range of variation for y is c-a to c+a (because $\sin bx$ varies from -1 to 1) and the question tells us the ropes height is from 1 to 5 feet, we have $3 - a =1$ and $3 + a = 5$. This gives us $a = 2$

$2\sin bx + 3$ has the same period as $\sin bx$. Now since the period of a sin function is $2\pi$, we are forced to choose $b = \frac12$.

Thus the final equation is
$y = 2\sin \frac{x}{2} + 3$

Side Note:Now since $4\pi$ is the period, it means $a\sin bx + c = a\sin b(x+4\pi) + c \Rightarrow bx+4\pi b = n\pi+ (-1)^n bx$. Clearly for n=2, the equation reads $4\pi b = 2\pi$ and thus b=2

6. Thank you very much Isomorphism!

I get it now. Finally. =)