# Horizontal range of 2 different angles

1. Jan 27, 2014

### negation

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

Show that, for a given initial speed, te horizontal range of a projectile is the same for launch angles $45° +\alpha$ and $45° - \alpha$

3. The attempt at a solution

$x1 = vi cos (45° + \alpha)$
$x2 = vi cos (45° - \alpha)$

tcomplete trajectory = $2 vi sin(45°\pm\alpha)/g$

Substituting tcomplete trajectory into x1 and x2:

$x1$ = $vi^2 sin (2(45°+\alpha))/g$
$x2$ = $vi^2 sin (2(45°-\alpha))/g$

2. Jan 27, 2014

### voko

So, are $x_1$ and $x_2$ equal?

3. Jan 27, 2014

### negation

It doesn't appears so.

4. Jan 27, 2014

### voko

Why not? Why can't $\sin (2(45^\circ + \alpha))$ be equal to $\sin (2(45^\circ - \alpha))$?

5. Jan 27, 2014

### negation

Let (45 + a) =x and (45 -a) = y
How does sin(2x) = sin(2y)?

6. Jan 27, 2014

### voko

You are trying to solve a harder problem that way. Stick with the original form.

7. Jan 27, 2014

### negation

The solution in the op is the answer?

8. Jan 27, 2014

### voko

It is very close to the answer. You just need to explain why those two sines are equal. You may want to review the trig identities that you know, or just go straight to the unit circle.

9. Jan 27, 2014

### negation

Ok. I shall further that attempt. It's probably in the form 2cos(x)sin(x)

10. Jan 27, 2014

### voko

One useful identity is $$\sin \left( \frac \pi 2 + \theta \right) = ?$$

11. Jan 27, 2014

### negation

Solved and Thanks!