Ratio of flight times at angles 45degree +/- alpha

[negation]

Homework Statement

Consider two projectiles launched on level ground with the same speed,
at $angles 45°\pm \alpha$.
Show that the ratio of their flight times is $tan 45°\pm \alpha$.

The Attempt at a Solution

projectile 1 launched at $Θ = 45° + \alpha$:

v_yf1 = v_yi1 + gt
t_apex1 = v_yi/g
t_{total flight time1} = 2v_yi1/g
= 2 vi sin (45 ° + $\alpha$)


projectile 2 launched at $Θ = 45° - \alpha$:

v_yf2 = v_yi2 + gt
t_apex2 = v_yi2/g
t_{total flight time2} = 2v_yi2/g
= = 2 vi sin (45 ° - $\alpha$)

 

[BvU]

Correct. Now write out the trigonometry!

 

[negation]

Correct. Now write out the trigonometry!

I'm not sure it's correct. I mean, intuitively, tan = sin / cos.

Both of my equation for the flight time are in sin.

 

[negation]

Correct. Now write out the trigonometry!

What do you mean by writing out the trigonometry?

 

[Chestermiller]

Incidentally, the given answer is incorrect. It should be $(tan(45) \pm2α)=1\pm2α$, and this only applies in the limit of small values of α.

You first take the ratio of the two flight times, and then apply the trig formulas for the sum and difference of two angles. Then you linearize with respect to α.

 

[gneill]

Incidentally, the given answer is incorrect. It should be $(tan(45) \pm2α)=1\pm2α$, and this only applies in the limit of small values of α.

You first take the ratio of the two flight times, and then apply the trig formulas for the sum and difference of two angles. Then you linearize with respect to α.

I don't think that the given answer is incorrect. $tan(\frac{\pi}{4} + \alpha)$ looks okay to me...

 

[Chestermiller]

I don't think that the given answer is incorrect. $tan(\frac{\pi}{4} + \alpha)$ looks okay to me...

Wow. Very cute. I must admit, I didn't think of putting the cosine of the complement in the denominator. Well, at least I got the right answer in the limit of small α. Not very satisfying for me, however.

Chet

 

[BvU]

Gentlemen, gentlemen: tan45°±α means tan (45°±α). Of course (right, negation?). And it is not a linearization, approximation or anything. It has to do with the funny trigonometric properties of $\frac{\pi}{4}$. Write it out !

I'm not sure it's correct. I mean, intuitively, tan = sin / cos.

Both of my equation for the flight time are in sin.

Your are correct, but: intuition? If you don't believe it, look it up (or make a drawing ;-) )

Your equations are in sin, yes, they are. The ratio of the flight times is sin (45 ° + α) / sin (45 ° - α). That are sines of sum and difference, which can be written out (as the hint suggested). Something physicists and Excel users look down on, but it can be done.

Here we go: using

$\sin(\frac{\pi}{4}+\alpha) = \sin(\frac{\pi}{4}) \cos\alpha + \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha + \sin\alpha \right )$ and

$\sin(\frac{\pi}{4}-\alpha) = \sin(\frac{\pi}{4}) \cos\alpha - \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha - \sin\alpha \right )$ to get a ratio

$\frac{ \cos\alpha + \sin\alpha}{ \cos\alpha - \sin\alpha}$

(sorry, it's been over 25 years since I last used TeX and it's a little rusty now, don't know how to get it in comparable font sizes any more)

Anyway, now divide by $\cos\alpha$ above and below to get a ratio

$\frac{ 1+ \tan\alpha}{ 1- \tan\alpha} =\frac{ \tan\frac{\pi}{4} + \tan\alpha}{ 1-\tan\frac{\pi}{4} \tan\alpha}$

If we remember (or look up, as I had to do)

$\tan(\gamma+\alpha) = \frac{\tan\gamma + \tan\alpha} {1- \tan\gamma \tan\alpha}$ we see that they are the same !

 

[Chestermiller]

Gentlemen, gentlemen: tan45°±α means tan (45°±α). Of course (right, negation?). And it is not a linearization, approximation or anything. It has to do with the funny trigonometric properties of $\frac{\pi}{4}$. Write it out !



Your are correct, but: intuition? If you don't believe it, look it up (or make a drawing ;-) )

Your equations are in sin, yes, they are. The ratio of the flight times is sin (45 ° + α) / sin (45 ° - α). That are sines of sum and difference, which can be written out (as the hint suggested). Something physicists and Excel users look down on, but it can be done.

Here we go: using

$\sin(\frac{\pi}{4}+\alpha) = \sin(\frac{\pi}{4}) \cos\alpha + \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha + \sin\alpha \right )$ and

$\sin(\frac{\pi}{4}-\alpha) = \sin(\frac{\pi}{4}) \cos\alpha - \cos(\frac{\pi}{4}) \sin\alpha =\frac{1}{2}\sqrt{2} \left ( \cos\alpha - \sin\alpha \right )$ to get a ratio

$\frac{ \cos\alpha + \sin\alpha}{ \cos\alpha - \sin\alpha}$

(sorry, it's been over 25 years since I last used TeX and it's a little rusty now, don't know how to get it in comparable font sizes any more)

Anyway, now divide by $\cos\alpha$ above and below to get a ratio

$\frac{ 1+ \tan\alpha}{ 1- \tan\alpha} =\frac{ \tan\frac{\pi}{4} + \tan\alpha}{ 1-\tan\frac{\pi}{4} \tan\alpha}$

If we remember (or look up, as I had to do)

$\tan(\gamma+\alpha) = \frac{\tan\gamma + \tan\alpha} {1- \tan\gamma \tan\alpha}$ we see that they are the same !

Gneill implied something much more elegant:

$$\frac{\sin(\frac{π}{4}+α)}{\sin(\frac{π}{4}-α)}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{2}-(\frac{π}{4}-α))}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{4}+α)}=\tan(\frac{π}{4}+α)$$

 

[BvU]

I love it! Missed the implication because I had it more or less worked out last night and was too sleepy to catch that. Anyway, dusting off old sum rules was a good thing.
Working back from the answer would have increased the chance of finding this very short proof...
And doodling with a plot of sin(x) and cos(x) probably also would have helped...

 

[gneill]

I love it! Missed the implication because I had it more or less worked out last night and was too sleepy to catch that. Anyway, dusting off old sum rules was a good thing.
Working back from the answer would have increased the chance of finding this very short proof...
And doodling with a plot of sin(x) and cos(x) probably also would have helped...

Good advice! Problem solving epiphanies often spring from dredging up the basic trig function relationships and identities. All those courses one thought done and dusted in the prerequisites course list regularly turn out to be a goldmine for these "aha!" moments.

 

[BvU]

I'm not sure negation has a satisfied feeling about all this help going over his head. How is it ?

 

[negation]

Gneill implied something much more elegant:

$$\frac{\sin(\frac{π}{4}+α)}{\sin(\frac{π}{4}-α)}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{2}-(\frac{π}{4}-α))}=\frac{\sin(\frac{π}{4}+α)}{\cos(\frac{π}{4}+α)}=\tan(\frac{π}{4}+α)$$

I must admit to the elegance and simplicity of the above equation. It's a good refresher. I had the time of flight in symbolic form for both particles fired at an arbitrary deviation of +/- alpha but couldn't reduce the ratio to the intended identity.

 

[negation]

I'm not sure negation has a satisfied feeling about all this help going over his head. How is it ?

I just got home from work. I'll be looking it through after taking a shower.

 

[negation]

Anyway, now divide by $\cos\alpha$ above and below to get a ratio

$\frac{ 1+ \tan\alpha}{ 1- \tan\alpha} =\frac{ \tan\frac{\pi}{4} + \tan\alpha}{ 1-\tan\frac{\pi}{4} \tan\alpha}$

Can I not multiply by cos(a) + sin(a)?

 

[BvU]

Would it help ?

 

[negation]

Would it help ?

You're right. It's not an appropriate step to arriving at the intended identity.

 

[negation]

You're right. It's not an appropriate step to arriving at the intended identity.

I understand how 1+ tan(a) is tautological to tan(pi/4) + tan(a)
but how does 1 - tan(a) = 1 - tan(pi/4) tan (a)?

 

[BvU]

tan(pi/4) = 1, so you can replace a 1 with this guy anywhere you want...

 

[negation]

tan(pi/4) = 1, so you can replace a 1 with this guy anywhere you want...

I know and which I did.

However, 1 + tan (a) : 1 - tan(a) = tan (pi/4 + a) : tan (pi/4 - a)

gives me [tan(pi/4) + tan(a)]/1 - tan(pi/4)tan(a) : [tan(pi/4) - tan(a)]/1 + tan(pi/4)tan(a)

I'm unable to reduce it further.

 

[BvU]

However, 1 + tan (a) : 1 - tan(a) = tan (pi/4 + a) : tan (pi/4 - a)

This is new. Where did that = come from ?

I remember writing

$\frac{ 1+ \tan\alpha}{ 1- \tan\alpha} =\frac{ \tan\frac{\pi}{4} + \tan\alpha}{ 1-\tan\frac{\pi}{4} \tan\alpha}$

which was exactly the expression for $\tan(\gamma+\alpha)$ with $\gamma=\frac{\pi}{4}$.

And the $\pm$ only comes in because for the ratio of a and b one can write a/b but also b/a.
I leave it to you to deduce that $\tan(\frac{\pi}{4}+\alpha)$ = 1/$\tan(\frac{\pi}{4}-\alpha)$

 

[negation]

This is new. Where did that = come from ?

I remember writing


which was exactly the expression for $\tan(\gamma+\alpha)$ with $\gamma=\frac{\pi}{4}$.

And the $\pm$ only comes in because for the ratio of a and b one can write a/b but also b/a.
I leave it to you to deduce that $\tan(\frac{\pi}{4}+\alpha)$ = 1/$\tan(\frac{\pi}{4}-\alpha)$

tan(pi/4 + a) = [tan(pi/4) + tan(a)]/[1 - tan(pi/4) tan(a)] This is understandable.
But from your post you equated [1 + tan(x)]/[1 - tan(x)] = [1 - tan(pi/4) tan(a)], how does
[1 - tan(x)] = [1 - tan(pi/4) tan(a)] follow from [1 + tan(x)]/[1 - tan(x)]?

 

[negation]

This is new. Where did that = come from ?

I remember writing


which was exactly the expression for $\tan(\gamma+\alpha)$ with $\gamma=\frac{\pi}{4}$.

And the $\pm$ only comes in because for the ratio of a and b one can write a/b but also b/a.
I leave it to you to deduce that $\tan(\frac{\pi}{4}+\alpha)$ = 1/$\tan(\frac{\pi}{4}-\alpha)$

Got it.