# 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