# Why angle of projectile has 2 solutions?

1. May 24, 2013

### e-zero

I have the final answer of:

sin2(theta) = 0.871

why does (theta) = 30.3 deg OR (theta) = 59.7 deg

I get why this could be physically, but I'm not sure why mathematically?

2. May 24, 2013

### Staff: Mentor

Mathematically the sqrt of sin^2(theta) would be sin(theta) or -sin(theta) giving two possible thetas as a solution in the domain of thetas from 0 to 90 degrees.

3. May 24, 2013

### jbriggs444

I believe that you are trying to solve "sin ( 2 theta ) = 0.871" for theta.

This yields two solutions because theta can be about 30 degrees and then 2 theta is about 60 degrees (high in the first quadrant) or theta can be about 60 degrees and then 2 theta is about 120 degrees (high in the second quadrant).

The arc-sine function on your calculator would, of course, return only the first of these two possibilities, ignoring the second quadrant solution.

4. May 24, 2013

### Aero_UoP

If you are solving the equation sin2θ=0.871 then, as jedishrfu said, sinθ = ± 0.933 which gives you a positive value for θ if the projectile is launched to go up from the axis of reference or a negative value of θ if the projectile is launched to go down.

5. May 24, 2013

### jbriggs444

One reason for discarding that interpretation is that the two solutions given in the original post were 30.3 and 59.7 degrees.

sin(2*30.3) = 0.871
sin(2*59.7) = 0.871

By contrast, if sin theta = ±0.933 then theta would be ±68.9 degrees. Those aren't even close to the solutions we're being asked to explain.

6. May 24, 2013

### Aero_UoP

First of all he hasn't made clear which equation he is solving. Is he solving the sin2θ=0.871 or the sin(2θ)=0.871? I covered just one of the 2 occasions.

Secondly, we cannot be sure that he has done the math correctly.

7. May 24, 2013

### Aero_UoP

Moreover, we don't even know what problem he is trying to solve. He just gave what he claims to be his final answer. We don't know anything about the rest of the solution.

8. May 26, 2013

### Redbelly98

Staff Emeritus
Well, the given solutions of 30.3 and 59.7 degrees are consistent with $\sin(2 \theta)$, so I think it is pretty clear which equation it is.

9. May 26, 2013

### e-zero

The final answer was derived from the formula:

R = (v0^2 * sin2θ) / g

where θ is the angle I am trying to solve for (in my case it is a football being kicked), R is the horizontal range, v0 is the initial velocity, and g is gravity.

I guess a another way to put the question is:

If I see sin2θ in a formula, does that mean that there is potentially two angles for the answer??

10. May 26, 2013

### Staff: Mentor

Depends on the constraints of theta for the problem at hand. In the current problem theta could range from 0 to 90 degrees. In the domain of sin (theta) from 0 to 180 degrees (ie positive sin value) there are two angles that have the same sin.

When you consider 2*theta and with a positive sin constraint you still have two angles within the range of 0 to 90 degrees.

11. May 27, 2013

### e-zero

So lets say for arguments sake that I had:

tan2(theta) = 0.871

What could be the solutions for theta?

12. May 27, 2013

### nasu

13. May 27, 2013

### e-zero

So mathematically speaking, what two angles are solutions for tan2(theta) = 0.871 ?

14. May 27, 2013

### Redbelly98

Staff Emeritus
You know one solution is
2θ=arctan(0.871)​
Then you have to think about whether the tangent function repeats any values during a period, the way sine and cosine do, which would generate additional solutions. Tangent does not do this, so we're good there.
Finally, we can add or subtract integer multiples of the tangent function's period, namely π, to the above equation,
2θ=arctan(0.871) ± nπ,​
and you check whether that generates additional solutions.

Discard solutions that differ by 2π from another solution, since they are really the same angle. Also discard any solution that is unphysical or for any reason meaningless. For example, a projectile launch angle should not be greater than 90°, unless you want to consider the projectile being fired backward for some reason, then it should be under 180°. On the other hand a compass heading could be anywhere from 0 to 360°.

For this last part, you are not relying on memorizing any rule. You have to think about the physical meaning behind the answers that the math spits out.

Last edited: May 27, 2013
15. May 27, 2013

### e-zero

I kinda a get what your saying...

When I go back to sin2(theta) = 0.871 I get the following:

http://www.wolframalpha.com/input/?i=sin(2*+x)=0.871

My question now is how do I know that the two angles are approximately 30.3 deg or 59.7 deg by looking at the graph in the above link?

16. May 27, 2013

### Staff: Mentor

The x axis is in radians so convert the 30.3 degrees radians and check again.

Remember 90 degrees is PI/2

17. May 28, 2013

### e-zero

Ok, can you explain how to appropriately calculate the solutions just using a calculator for the following?

sin2θ = 0.871
cos2θ = 0.871
tan2θ = 0.871

18. May 28, 2013

### jbriggs444

If sin 2θ = 0.871 then...

2θ = arcsin ( 0.871 ) or, since the sin function is symmetric about 90 degrees...
2θ = 180 - arcsin ( 0.871 )

So

θ = arcsin ( 0.871 ) / 2 or
θ = ( 180 - arcsin ( 0.871 ) ) / 2

So put your calculator into Degrees mode (often the default).
Key in 0.871
Click on inverse sin -- giving a figure of 60.575
Click on the divide by sign and key in 2.
Click on the equal sign -- giving a figure of 30.3

Key in 180 and press the minus sign
Key in 0.871 and click on inverse sin -- giving a figure of 60.575
Click on the equal sign -- giving a figure of 119.425 [this is 180 - arcsin(0.871)]
Click on the divide by sign and key in 2.
Click on the equal sign -- giving a figure of 59.7

19. May 28, 2013

### e-zero

So you do the same for cos and tan?

20. May 28, 2013

### jbriggs444

21. May 28, 2013

### Redbelly98

Staff Emeritus
That gets us 2 of the 4 solutions, if we are solving this as a math problem. We need to add integer multiples of the period 2π to those equations to find the other solutions.

2θ = arcsin ( 0.871 )+2π
gives another solution. Do the same for the "180-arcsin(0.871)" equation, and you get yet another solution.

Adding twice the period, 2*2π, gets us 390.3°, which is equivalent to 30.3° and thus not a distinct solution, so we can stop there.

22. May 28, 2013

### e-zero

Ok, at this point I'm just trying to understand the trigonometry here (just the math).

So your saying that sin2 ø=0.871 has 4 solutions?

Do you have a link that can explain this further? Or can you explain how to find all solutions for a given trig function? (currently searching google...

23. May 29, 2013

### Redbelly98

Staff Emeritus
I tried to explain it in Post #21.

Alternatively you can take the graph you got at wolfram, then extend the sine graph for a full 2*pi, and see that it intersects the y=0.871 line in two more places.

EDIT:
If you can get a pre-calc textbook, you can look at the section on solving trig equations. Or try googling "solving trig equations". Here's is one of the links I found from a google search:
Pages 15-17 have an example that is similar to what we have been discussing.

Last edited: May 29, 2013
24. May 29, 2013

### DEvens

I'm guessing, since you give very little context.

The short answer is, because this function is periodic. There are two places where a sine wave has value 0.871 in each period.

In context: It seems (from the subject) like you are solving some kid of projectile problem. When you are firing a projectile at a target closer than the maximum distance, then there usually are two solutions. Neglecting air resistance and for the simplest projectile problem of same-level to same-level, the maximum distance applies at 45 degrees elevation. To shorten the distance you can either raise or lower the angle. Raising it means the projectile flies higher, with a slower forward speed. Lowering it means the projectile flies less high, with a faster forward speed.

25. May 29, 2013

### jbriggs444

Two obvious physical solutions correspond to shooting the projectile from the launch point toward the target.

The two additional solutions correspond to having the projectile come back from the target to align with the butt of a down-and-left-pointing cannon. The time of flight (measured from cannon to target) is negative for these additional two cases.

Mathematically, as RedBelly98 has politely pointed out, the additional solutions are at +180 degrees (or +pi radians) from the obvious solutions. [Roughly 210 degrees and 240 degrees].