Horizontal range of 2 different angles

[negation]

Homework Statement

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

The Attempt at a Solution

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

t_{complete trajectory} = $2 vi sin(45°\pm\alpha)/g$

Substituting t_{complete trajectory} into x1 and x2:

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

 

[voko]

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

 

[negation]

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

It doesn't appears so.

 

[voko]

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

 

[negation]

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

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

 

[voko]

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

 

[negation]

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

The solution in the op is the answer?

 

[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.

 

[negation]

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.

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

 

[voko]

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

 

[negation]

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

Solved and Thanks!