MHB Trigonometric Simplification for Projectile Motion Equation

[Theia]

Hi!

Long ago (more than 5 years now, actually) I got stuck with a trigonometrig formula and I haven't been able to got the point. I had an equation (with respect to $$f$$):

$$(r - a)L^2 + r\tan \alpha \cos f \cdot L + a\tan^2 \alpha = 0$$,

where

$$L = \sin f - \tan \alpha \cos f$$.

According to my partially lost notes, some CAS has been simplified this into

$$\tfrac{1}{2} \sec ^2 \alpha \sin f [(2a - r)\sin (2\alpha - f) + r\sin f] = 0$$.

Now I'd like to ask you, is this true at all? And if it is true, what tricks(?) were used? Thank you so much! ^^

 

[I like Serena]

Hi!

Long ago (more than 5 years now, actually) I got stuck with a trigonometrig formula and I haven't been able to got the point. I had an equation (with respect to $$f$$):

$$(r - a)L^2 + r\tan \alpha \cos f \cdot L + a\tan^2 \alpha = 0$$,

where

$$L = \sin f - \tan \alpha \cos f$$.

According to my partially lost notes, some CAS has been simplified this into

$$\tfrac{1}{2} \sec ^2 \alpha \sin f [(2a - r)\sin (2\alpha - f) + r\sin f] = 0$$.

Now I'd like to ask you, is this true at all? And if it is true, what tricks(?) were used? Thank you so much! ^^

Hey Theia! ;)

Let's see...
$$(r - a)(\sin f - \tan α \cos f)^2 + r\tan α \cos f \cdot (\sin f - \tan α \cos f) + a\tan^2 α = 0 \\
(r-a)(\sin^2 f - 2\tan α \sin f\cos f + \tan^2α \cos^2 f) + r\tan α \cos f \sin f - r\tan^2 α \cos^2 f + a\tan^2 α = 0 \\
(2a-r)\tan α \cos f \sin f + a(-\sin^2 f - \tan^2α \cos^2 f + \tan^2 α)+ r\sin^2 f = 0 \\
(2a-r)\tan α \cos f \sin f + a(-\sin^2 f + \tan^2 α(1 - \cos^2 f )) + r\sin^2 f = 0 \\
(2a-r)\tan α \cos f \sin f - a (1 - \tan^2 α)\sin^2 f + r\sin^2 f = 0 \\
\frac 12 \sec^2 α \sin f\Big[(2a-r)2\sin α \cos α\cos f - 2a (\cos^2 α - \sin^2 α)\sin f + 2r\cos^2 α\sin f \Big] = 0 \\
\frac 12 \sec^2 α \sin f\Big[(2a-r)(2\sin α \cos α\cos f - (\cos^2 α - \sin^2 α)\sin f) - r (\cos^2 α - \sin^2 α)\sin f + 2r\cos^2 α\sin f \Big] = 0 \\
\frac 12 \sec^2 α \sin f\Big[(2a-r)(\sin 2α \cos f - \cos 2α\sin f) + r (\cos^2 α + \sin^2 α)\sin f \Big] = 0 \\
\frac 12 \sec^2 α \sin f\Big[(2a-r)\sin(2α - f) + r \sin f \Big] = 0 \\
\sin f = 0 \quad\vee\quad (2a-r)\sin(2α - f) + r \sin f = 0 \\
$$
The "tricks" that we've used are:
$$
\cos^2 α + \sin^2 α = 1 \\
\sin 2α = 2\sin α\cos α \\
\cos 2α = \cos^2 α - \sin^2 α \\
\sin(α-β)=\sin α\cos β - \cos α \sin β
$$

 

[Theia]

Wow! Thank you so much! It looks like the key was some not so obvious manipulations to the equation. *wonders why it is so difficult to add 0 sometimes... :D *

As a side note, this equation was part of my study how to describe projectile motion under Newtonian $$-\gamma \tfrac{Mm}{r^2}$$ gravity (e.g. on Moon) without messing too much with celestial mechanics. Everything else went smoothly, except this simplification... But once the numerics worked, I didn't pay attention to it until now. :D Thank you again!