Proving Sin(120) = Sin(60) with Trigonometry

[simphys]

Homework Statement

Please refer to the picture (not really needed) so that you are able to see in what context it is meant. —> question is under the figure

Relevant Equations

R=v^2sin^2(2alpha) / g

It is about that the rznge of 60 degrees = R of 30 degrees, but how would I prove that?

Sin(120) needs to equal sin(60)
How can i prove that theyll be the same range(without air resistance?)

My take: (only looking at the sin(alpha) part as that neefs to be equal)

using trig identity
- Sin(120) = sin(90+30) = sin90cos30 + 0 = cos 30
- Sin(60) = cos30 compelemntary

So they’ll be equal.

Not really comfortable with complementary anglesso wanted to assure that this is correct. TIA!

 

[Lnewqban]

I believe the relevant equation is incorrect, the way is shown in the OP.
Please, see:
https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/

I would draw those two double angles (2x30 and 2x60) and see what you can deduce at a glace regarding the value of their sines for that particular combination of launch angles (30 and 60).

 

[BvU]

Not really comfortable with complementary angles so wanted to assure that this is correct

As pointed out, the expression for $R$ contains a sine, not sin^2.

Solving $R_1 = R_2$ with same $v$ and $g$ ends up with an equation $\sin 2\alpha_1 = \sin 2\alpha_2$. Draw a plot of $\sin 2\alpha$ for $\ [0,{\pi\over 2}]\$ to see the relationship between $\alpha_1$ and $\alpha_2$.

$\$

 

[simphys]

I believe the relevant equation is incorrect, the way is shown in the OP.
Please, see:
https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/

I would draw those two double angles (2x30 and 2x60) and see what you can deduce at a glace regarding the value of their sines for that particular combination of launch angles (30 and 60).

Thanks for the help! and indeed, it was a typo, my apologies.

As pointed out, the expression for $R$ contains a sine, not sin^2.

Solving $R_1 = R_2$ with same $v$ and $g$ ends up with an equation $\sin 2\alpha_1 = \sin 2\alpha_2$. Draw a plot of $\sin 2\alpha$ for $\ [0,{\pi\over 2}]\$ to see the relationship between $\alpha_1$ and $\alpha_2$.

$\$

Thanks for the help as well! (and that was indeed a type my apologies)You guys said to draw, I didn't think of it at all, was solving it algebraically, thanks for the tip!