Solve for Theta: How to Divide Sin?

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In summary: I know but I didn't know what to do, can you help me?If you have sinx = A, then x = sin-1(A).f is a function. What happened to it? You can't just ignore it.Thanks, but please bear with me, my goal is to understand everything in mathematics as much as I can. I hope I am not being a bother but I am not satisfied with knowing that I should simply multiply by the inverse on both sides.
  • #1
Nano-Passion
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Homework Statement


Its a word problem but I will just state everything here in simple form.

Given
------------
V initial = 100ft/s
r=300ft
r= 1/32 V^2 Sin 2theta

Unknown
---------
Solve for theta

The Attempt at a Solution


300=1/32 (100)^2 sin 2theta
300= 1/32 (10000) sin 2theta
300= 312.5 sin 2theta
300/312.5= sin 2theta
.96 = sin 2theta

I know the next step would be to divide by sin but how?
 
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  • #2
Nano-Passion said:

Homework Statement


Its a word problem but I will just state everything here in simple form.

Given
------------
V initial = 100ft/s
r=300ft
r= 1/32 V^2 Sin 2theta

Unknown
---------
Solve for theta

The Attempt at a Solution


300=1/32 (100)^2 sin 2theta
300= 1/32 (10000) sin 2theta
300= 312.5 sin 2theta
300/312.5= sin 2theta
.96 = sin 2theta

I know the next step would be to divide by sin but how?
"sin" is not a number, and is not multiplying 2theta. Sine is a function, so your line at the end is similar to .96 = f(2x).

If you have a variable that is the argument to a function, what can you do to get at the function's argument?
 
  • #3
You have 2θ there. That is really like saying Sin(θ+θ) isn't it?
 
  • #4
(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
Nano-Passion said:
.96 = sin 2theta

I know the next step would be to divide by sin but how?

nooo … the next step is to use sine tables (or the arcsin button on your calculator) :smile:
 
  • #5
Mark44 said:
"sin" is not a number, and is not multiplying 2theta. Sine is a function, so your line at the end is similar to .96 = f(2x).

If you have a variable that is the argument to a function, what can you do to get at the function's argument?

hmm..

f(2x)=.96
2x=.96
x=.48? -.-grrr

I'm having a little trouble =/ I can't seem to get the inverse ..

What I did before was I asked myself:
sin of what? = .96 <-- plugged it in calculator as sin of inverse =/
2θ = 1.287
θ = .6435

But I want to figure our what's going on in between.
tiny-tim said:
(have a theta: θ and try using the X2 tag just above the Reply box :wink:)nooo … the next step is to use sine tables (or the arcsin button on your calculator) :smile:

Thanks but I want to really understand the steps in between and not just get the answer through the calculator.

And is there a button here for theta?
 
  • #6
Nano-Passion said:
hmm..

f(2x)=.96
2x=.96
f is a function. What happened to it? You can't just ignore it.
Nano-Passion said:
x=.48? -.-grrr

I'm having a little trouble =/ I can't seem to get the inverse ..



Thanks but I want to really understand the steps in between and not just get the answer through the calculator.

And is there a button here for theta?
 
  • #7
If you have sinx = A, then x = sin-1(A).
 
  • #8
Mark44 said:
f is a function. What happened to it? You can't just ignore it.

I know but I didn't know what to do, can you help me?

I'm used to things such as:
f(x) = 2x+4
y = 2x+4
y/2 -2 = x
f-1x = 1/2y - 2

but when it comes to

f(2x) = .96 then I'm confused on what to do.
 
  • #9
Isn't sin2[tex]\theta[/tex] a trigonometric Identity that becomes 2 sin [tex]\theta[/tex] cos[tex]\theta[/tex]?
 
  • #10
In your problem, the function is the sine function, and your equation is:
sin(2x) = .96

(I'm using x instead of theta.)
The thing to do is to apply the inverse sin function to both sides.

sin-1(sin(2x)) = sin-1(.96)

==> 2x = sin-1(.96)
==> x = (1/2)sin-1(.96)

This will give you one value for x, but it might be that your problem calls for other solutions. If so, you will need to use some of the ideas from trig to get the other solutions.
 
  • #11
AwesomeSN said:
Isn't sin2[tex]\theta[/tex] a trigonometric Identity that becomes 2 sin [tex]\theta[/tex] cos[tex]\theta[/tex]?

But in this problem, that's the wrong way to go. Making this replacement turns the problem into [itex]2 sin \theta cos\theta = .96[/itex]

and this doesn't get you closer to a solution. Using inverse functions does.
 
  • #12
Mark44 said:
In your problem, the function is the sine function, and your equation is:
sin(2x) = .96

(I'm using x instead of theta.)
The thing to do is to apply the inverse sin function to both sides.

sin-1(sin(2x)) = sin-1(.96)

==> 2x = sin-1(.96)
==> x = (1/2)sin-1(.96)

This will give you one value for x, but it might be that your problem calls for other solutions. If so, you will need to use some of the ideas from trig to get the other solutions.

Thanks, but please bare with me, my goal is to understand everything in mathematics as much as I can. I hope I am not being a bother but I am not satisfied with knowing that I should simply multiply by the inverse on both sides.

My question is, how does sin-1 cancel with sin when you multiply them together? It doesn't sound mathematically logical? I need the steps in between. :blushing:
 
  • #13
Nano-Passion said:
Thanks, but please bare with me, my goal is to understand everything in mathematics as much as I can. I hope I am not being a bother but I am not satisfied with knowing that I should simply multiply by the inverse on both sides.
No, you're not being a bother. If you don't understand something it's better to keep asking questions until things are clear.

One thing that you need to understand here is that the operation is NOT multiplication. This is something you have not been clear on since your first post in this thread.

sin(2x) is sometimes written as sin 2x. In either form it is NOT sin times 2x - it's sin of 2x. Similar to what I wrote earlier - f(2x) is not f times 2x. It's f OF 2x, where f was the name of some unspecified function.
Nano-Passion said:
My question is, how does sin-1 cancel with sin when you multiply them together?
They're not being multiplied. What is happening is that I am forming a composite function. Whenever you have a function that has an inverse, applying them together in either order gives you the identity function, the function that leaves its argument completely unchanged.

For example, if f(x) = 2x + 3, then the inverse of this function is f-1(x) = (x - 3)/2.

f(f-1(2) = 2 and f-1(f(0)) = 0.
You can verify these statements by using the formulas for the functions.

In a similar way sin(sin-1(x) = x, but there are some restrictions of the values of x that are allowed. Also, in the opposite order, sin-1(sin(x)) = x, and there are some restrictions here, as well. In the first equation, x has to be between -1 and +1, inclusive. In the second equation, x has to be between -pi/2 and +pi/2 if you're working in radians, or between -90 deg and +90 deg, if you're working in degrees.
Nano-Passion said:
It doesn't sound mathematically logical? I need the steps in between. :blushing:
 
  • #14
Yes, sin-1(sin(theta) ) = theta
 
  • #15
RocketSci5KN said:
Yes, sin-1(sin(theta) ) = theta

That's not always true. Did you not read Mark44's post above?

For instance, what is
[tex]\sin^{-1} (\sin 3\pi/4)[/tex]
? Hint: it's not 3π/4.
 
  • #16
RocketSci5KN said:
Yes, sin-1(sin(theta) ) = theta
Noooooooo! :frown: You have to apply methods for solving trigonometric equations! Pretending it's an ordinary invertible operation you can simply "undo" is incorrect!
 
  • #17
Do they still teach English students how to diagram sentences? If you've seen that, there's a similar thing for mathematical expressions.

The diagram (which can called a "parse tree") for the expression 2x+1 would be:
Code:
     +
    / \
   /   \
  *     1
 / \
2   x
Note this reflects how it's computed -- if you substitute x=4 and computed, you would
  • Replace x with 4
  • Compute 2*4 and replace that little part of the tree with the result:
    Code:
         +
        / \
       /   \
      8     1
  • Compute 8+1, and replace the tree with 9


Functions, such as sin are like + and * and other operations. The diagram for the expression
sin 2x​
looks like
Code:
 sin
  |
  *
 / \
2   x
And for
arcsin sin 2x​
it would be
Code:
arcsin
  |
 sin
  |
  *
 / \
2   x

I hope this helps you understand how to read mathematical expressions.
 
  • #18
Oops... forgot about the range of theta this IS true for, as noted by Mark44.
 
  • #19
Hurkyl said:
Do they still teach English students how to diagram sentences?
AFAIK, they don't, and more's the pity. It's been out of the curriculum in the U.S. for so long, that I would guess that most high school English teachers here have never been exposed to this concept.
 
  • #20
Mark44 said:
AFAIK, they don't, and more's the pity. It's been out of the curriculum in the U.S. for so long, that I would guess that most high school English teachers here have never been exposed to this concept.

Really? They still do it in our school.:bugeye:
 
  • #21
Nano-Passion said:
Thanks, but please bare with me, my goal is to understand everything in mathematics as much as I can. I hope I am not being a bother but I am not satisfied with knowing that I should simply multiply by the inverse on both sides.

My question is, how does sin-1 cancel with sin when you multiply them together? It doesn't sound mathematically logical? I need the steps in between. :blushing:

Let w = 2*x, so you have the equation sin(w) = 0.96. Once you have found w it is easy to get x.

So, look at your equation sin(w) = v (where v happens to be 0.96 in your case, but in some other problem might have a different value). Imagine looking at the plot of y = sin(w) for w between -pi/2 and + pi/2 (measuring angles in *radians*). The y-values in the plot go from y = -1 at w = -pi/2 up to y = +1 at w = pi/2, and the curve is increasing as w increases from -pi/2 to + pi/2. Now, for a value v strictly between -1 and +1 there will be a single value of w that solves the equation sin(w) = v; that will occur at the intersection between the graph y = sin(w) and the horizontal line at y = v. The value of w where that occurs is called the "inverse sine of v", or sin^(-1)(v); it is just the solution of the equation sin(w) = v.

We don't multiply sin and sin^(-1); rather, we "compose" them, in the sense that
sin[sin^(-1)(v)] = v and sin(-1)[sin(w)] = w.

RGV
 
  • #22
[itex]sin^{-1}(sin(x))= x[/itex] is not correct because sine is not a "one to one" function: [itex]sin(x)= sin(\pi- x)= sin(2\pi+ x)[/itex] etc.

If you have sin(2x)= .93, then you can use your calculator to find one value of 2x (there is no simple way to compute the inverse sine with paper and pencil- that's why people are telling you to use your calculator). subtract from [itex]\pi[/itex] and then add any multiple of [itex]2\pi[/itex] to both of those.
 

What is "Solve for Theta: How to Divide Sin"?

"Solve for Theta: How to Divide Sin" is a mathematical concept that involves finding the value of theta (θ) in a trigonometric equation where sinθ is being divided by another number or expression.

Why is it important to know how to divide sin?

Dividing sinθ is a common operation in trigonometry and is useful in solving various problems, such as finding the height of a building or the distance between two objects. Knowing how to divide sin allows for more accurate and efficient calculations.

What are the steps to solve for theta when dividing sin?

The steps to solve for theta when dividing sin are: 1) Identify the given equation and determine the value of the other trigonometric function (cos, tan, etc.) that is being divided by sinθ. 2) Use the reciprocal identity of sine (cscθ) to rewrite the equation. 3) Isolate sinθ on one side of the equation. 4) Take the inverse sine (arcsin) of both sides to solve for θ. 5) Check the solution by plugging it back into the original equation.

Can any number or expression be divided by sin?

No, only numbers or expressions that represent a ratio of sides in a right triangle can be divided by sin. This is because sin is a trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right triangle.

Are there any special cases when dividing sin?

Yes, when dividing sin by zero, the result is undefined. This is because division by zero is undefined in mathematics. Additionally, when dividing by negative values, the sign of the result will depend on the quadrant in which the angle θ lies.

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