# How to divide sin?

1. May 2, 2011

### Nano-Passion

1. The problem statement, all variables and given/known data
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

3. 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???

2. May 2, 2011

### Staff: Mentor

"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. May 2, 2011

### QuarkCharmer

You have 2θ there. That is really like saying Sin(θ+θ) isn't it?

4. May 2, 2011

### tiny-tim

(have a theta: θ and try using the X2 tag just above the Reply box )
nooo … the next step is to use sine tables (or the arcsin button on your calculator)

5. May 2, 2011

### Nano-Passion

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 whats going on in between.
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. May 2, 2011

### Staff: Mentor

f is a function. What happened to it? You can't just ignore it.

7. May 2, 2011

### rock.freak667

If you have sinx = A, then x = sin-1(A).

8. May 2, 2011

### Nano-Passion

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. May 2, 2011

### AwesomeSN

Isn't sin2$$\theta$$ a trigonometric Identity that becomes 2 sin $$\theta$$ cos$$\theta$$?

10. May 2, 2011

### Staff: Mentor

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. May 2, 2011

### Staff: Mentor

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

and this doesn't get you closer to a solution. Using inverse functions does.

12. May 2, 2011

### Nano-Passion

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.

13. May 2, 2011

### Staff: Mentor

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.
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.

14. May 2, 2011

### RocketSci5KN

Yes, sin-1(sin(theta) ) = theta

15. May 2, 2011

### eumyang

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

For instance, what is
$$\sin^{-1} (\sin 3\pi/4)$$
? Hint: it's not 3π/4.

16. May 2, 2011

### Hurkyl

Staff Emeritus
Noooooooo! You have to apply methods for solving trigonometric equations! Pretending it's an ordinary invertible operation you can simply "undo" is incorrect!

17. May 2, 2011

### Hurkyl

Staff Emeritus
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 (Text):

+
/ \
/   \
*     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 (Text):

+
/ \
/   \
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 (Text):
sin
|
*
/ \
2   x
And for
arcsin sin 2x​
it would be
Code (Text):
arcsin
|
sin
|
*
/ \
2   x
I hope this helps you understand how to read mathematical expressions.

18. May 2, 2011

### RocketSci5KN

Oops... forgot about the range of theta this IS true for, as noted by Mark44.

19. May 2, 2011

### Staff: Mentor

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. May 5, 2011

### BloodyFrozen

Really? They still do it in our school.