Question

help me understand (half angle formula, trigonometry) picture problem

Answer 1

Just rearrange the formula by taking the inverse sine of both sides. we get
\[\frac{\theta}{2}=\sin^{-1}\left(\frac{1}{M}\right)\] and hence
\[\theta=2\sin^{-1}\left(\frac{1}{M}\right)\]the answer is then
\[\theta=47.156^{o}\]

you can then find the distance behind the rocket the sonic cone reaches the ground from trigonometry. between the ground and the sonic cone is given by the angle calculated above. Therefore the horizontal distance x is given by \[x=\frac{3000}{\tan\theta}\]
hence x = 3235 ft.

Answer 2

I don't think we can just rearrange the formula like that cause sin x/2 = \[\pm \sqrt{1-\cos \Theta \div2}\]

Answer 3

er (1 - cos theta) / 2

Answer 4

Sorry my answer for part B is incorrect. the angle to use is theta/2, theta is the full angle. therefore the answer is x = 6874 ft.

Answer 5

its a half angle formula i dont think it can be split up

Answer 6

Ok lets use the identity of \[\sin(\theta/2)=\sqrt{\frac{1-cos\theta}{2}}\] we get the formula \[\frac{1-cos\theta}{2}=\left(\frac{1}{M}\right)^{2}\] rearrange, we get \[cos\theta=1-2\left(\frac{1}{M}\right)^{2}\]
take the inverse cosine and we get \[\theta=\cos^{-1}\left(1-\left(\frac{1}{M}\right)^{2}\right) \]

substitute M = 2.5 into it and you get the value of

\[\theta=47.156^{o}\]

using the half angle identity for sin(theta/2) only complicates the matter.

Answer 7

as you can see both methods are equivalent, and teh original method is easier. it may be a half angle, but you can take the inverse sine of it no problem, as there are no mathematical rules telling you cant.

Answer 8

i put 47 degrees and he marked it wrong

Answer 9

i guess he overlooked it

Answer 10

Was he perhaps looking for more precision? If not, then I am pretty certain the marker is is wrong. However, my wife is a professional mathematician, and I will alert her to this and see if she can shed more light on it.

Answer 11

well i take my test super sloppy i have math everywhere he circled something else when i drew an arrow from the math i got 47 degrees from

Answer 12

can you explain the second part some more please

Answer 13

oh so where the sonic cone and the ground touch is the same angle and the one of the sonic cone?

Answer 14

Consider the attached diagram. Ok, we now know the angle theta to be 47.156 degrees. But we know that the the cone is emitted symmetrically so we can bisect this angle by a horizontal line that is parallel to the ground, resulting in the angle between this line and the sonic cone to be theta/2. This also means that the angle between the ground and sonic cone will be theta/2. The angle between the height h and the sonic cone will be 90 - theta/2, because we have a right angle triangle. You can verify this a number of ways, but whatever way you are happiest with works best.

So choose an angle to work with, and solve using trigonometry. I chose the one between the ground and the cone. We know that tan(theta/2) = h/x. where x is the horizontal distance, so I just solved for x. if you took the other angle then we have sin(90-theta/2) = x/h and again solve for x, since we know that h = 3000 ft, and we have calculated theta from the first part.

Answer 15

crazy, symmetrical triangles I would of never thought of that

Answer 16

you are one smart fellow i applaud you hehe you have been a big help thanks a bunch

Answer 17

Hi I also get 47.156 degrees for theta and then 6,874 ft for the horizontal distance. The marker was probably wanting more precision, that is decimal places, to your answer for theta. Other than that, the marker may have just got annoyed with the maths being sloppy... they are human after all.

Just to clarify, the mathematics that is permitted, if \[\sin \left( \frac{\theta}{2} \right)=A\] then you are allowed to take the inverse sine of both sides and so you get \[\sin^{-1} \left( A \right)= \frac{\theta}{2}\]
However if you had \[\frac{\sin \theta}{3}=B\] then you would need to multiple both sides by 3 first, giving \[\sin \theta = 3 B\], and then you can take the inverse sine of both sides to give \[\theta = \sin^{-1} \left( 3 B \right)\]
I hope this helps :)

Answer 18

you guys are too much xD thanks i had no clue you can do that with a half angle

Answer 19

Thank you my lovely wife, who also pointed out the an error in an above post, where a 2 went missing. I wrote \[\theta=\cos^{-1} \left(1-\left(\frac{1}{M}\right)^2\right)\]when it should read\[\theta=\cos^{-1} \left(1-2\left(\frac{1}{M}\right)^2\right)\]

Answer 20

You are very welcome. Hope it helps.