projectiles question

Shox

i know this question probably seems retarded but its just getting on my nerves at this stage...:mad:

CosBCos(A-B)-SinBSin(A-B)

Does this not equal CosB

Yet the marking scheme i have says it equals Cos (2B-A)

Any help?

PurpleFistMixer

I'm getting it as CosA actually...

Let's see,

CosBCos(A-B)-SinBSin(A-B) =

(cosB)(cosAcosB + sinAsinB) - sinB(sinAcosB - cosAsinB)
cosAcos^2B + sinAsinBcosB - sinAsinBcosB + cosAsin^2B
(take out cos A, middle two terms cancel anyway)
cosA(cos^2B + sin^2B)
cosA(1)
= cosA.

Unless I'm getting the old cos(a-b) formulae wrong.. which is possible!

ZorbaTehZ

Just use the formula straight:
Let X = B, Y = (A-B)

Formula is :
Cos (X + Y) = Cos X . Cos Y - Sin X . Sin Y
. . .Cos( (B) + (A-B)) = Cos ( B-B + A) = Cos A

Yup PFM is right it's Cos A. What is the question shox?

JC 2K3

cosBcos(A-B) = (cosA + cos(2B-A))/2
sinBsin(A-B) = (cos(2B-A) - cosA)/2

therefore: cosBcos(A-B)-sinBsin(A-B)=
(cosA + cos(2B-A) - cos(2B-A) + cosA)/2
=cosA

Whoever wrote the marking scheme clearly used this method and mixed up the signs somewhere, probably writing sinBsin(A-B) = (cosA - cos(2B-A))/2 by mistake.

Shox

well its 2 questions im having the problem with.

2005 (b)

I got the range as being :(on marking scheme)
2u^2SinA(CosACosB-SinASinB)
gCos^2B


= 2u^2SinACos(A+B)
gCos^2B

But then when i do whrn range is max:

Cos(A+B)=0
A+B=90
A=90-B

But the answer for this has to be
A=45-B
2

So obviously in my range it needs to be Cos(2A+B)
But how do you get that??





2nd Prob 2003 (b)

For my range im getting:

u^2.2Sin(A-B)[Cos(A-B)CosB-Sin(A-B)SinB]
gCos^2B

u^2.2Sin(A-B).Cos(A-B+B)
gCos^2B

u^2.2Sin(A-B).CosA
gCos^2B

Answer at back is:

u^2[Sin(2A-B)-SinB]
gCos^2B

Im lost i can see they're using Sin2X=2SinXCosX

But in mine the X's above arent the same i
presume thats where my mistake is

ZorbaTehZ

2005 (b)

This a long part, so I will only write out the part you mentioned.
Finishing b, part one you get as the answer:

S(x) =
2u^2SinA(CosACosB-SinASinB)
gCos^2 B

Bit in the brackets changes to Cos(A-B) . . .from the Math Tables.
2u^2SinA(Cos(A-B))
gCos^2 B

2u^2 . [(SinA)(Cos(A-B))] . . .This is where you erred.
gCos^2 B

Only take the bit in the square brackets because...
You need to find the value of alpha that will yield greatest distance, so you can take everything else as being constant - nevertheless you do still need to take into account everything that does include alpha.
. . .(Sin A)(Cos(A-B))

Use the product to sum rule in the Math Tables that looks like 2SinACosB - and you get:
1/2 (Sin (A + (A-B)) + 1/2(Sin (A - (A-B))
. . .1/2(Sin (2A-B)) + 1/2(Sin

Disregard the 1/2(Sin as a constant and differentiate:
(Taking B as a constant)
ds(x)/dA [1/2 Sin (2A-B)] = 1/2 . Cos (2A-B) . 2
. . . = Cos(2A-B)

Maximum value at 0
Cos (2A-B) = 0
2A-B = 90
2A = 90 - B
A = 45 - B/2

ZorbaTehZ

2003 (b)
Use this rule instead:
2SinXCosY = Sin (X+Y) + Sin(X-Y)
. . .and let X= A-B and Y= A

. . .Sin(A-B+A) + Sin(A-B-A)
. . .Sin(2A-B) + Sin(-B)
. . .Sin(2A-B) - SinB

Shox

Thanks Zorba.:)

I see my mistake now.

Seems so obvious now after you've explained.