Check your question. As it is written, the first statement does not lead to the second. Here are the steps. For clarity, I'm going to say (F/B)^(μ*cotα) = Z, because it's a big messy term that appears in both statements. It won't get broken up, just moved around; it doesn't deserve more than one letter.
(F/B)^(μ*cotα) = (2μ*P*cotα + 2Y)/Y
Z = (2μ*P*cotα + 2Y)/Y
---------------------------------multiply by Y
YZ = 2μ*P*cotα + 2Y
---------------------------------subtract 2Y
YZ-2Y = 2μ*P*cotα
---------------------------------divide by 2μ and factor out Y
Y(Z-2)/2μ = P*cotα
---------------------------------cotα = 1/tanα, so multiply by tanα
Y(tanα)(Z-2)/2μ = P
---------------------------------now P is separated, as in the answer, and so is Y.
I'm going to take a sideways step now, to show why I don't think this is soluble. In a standard proof, I'd keep manipulating this P to look like the P I'm proving. Instead, I'm going to set them equal to each other, allowing me to cancel some stuff out and boil the problem down to its simplest elements. If it really is provable, then P always equals P and these two P equivalents should cancel out to 1 = 1.
P = Y(tanα)(Z-2)/2μ = Y(1+((tanα)/μ))(1-Z) ------ remember the Z substitution I made in the first step.
---------------------------------divide by Y
(tanα)(Z-2)/2μ = (1+((tanα)/μ))(1-Z)
---------------------------------there aren't any binomials the same between right and left (Z-2 and Z-1 are very different animals), but the right side binomials can be multiplied together.
(tanα)(Z-2)/2μ = 1 + (tanα)/μ - Z - Z((tanα)/μ)
Here it looks pretty nasty. Two terms added together on the left, four terms added together on the right, that's almost impossible to cancel out. But maybe expanding Z will help us. I'm going back a step to before the binomial expansion.
The trouble here is, even though Z has α and μ in it, they're in an exponent. They're not coming out of there. We can't even arrange the equation around far enough to try to simplify it with logarithms. We might as well keep calling it Z.
I can't move the problem past this point. In my opinion it's either it's miswritten or it's insoluble.
Edit: Dang, while I was working all this out, Michael went and worked out what the typo was. Curse him for competence!
hi chum this an uncomplicated question in line with linear equations you ought to use transposition approach in this technique while a term is transfered from one section to different its sign turns into opposite answer: x+4-7x=22 =4-7x=22-x (right here x comes from the left section its sign adjustments) =4=22-x+7x(right here 7x comes from the left section its sign adjustments) =4-22=6x(same) =-18=6x x=-18/6 =-3 Bye Bye!!!!!!!!!!!!!!!!!!!!!!
Comments
Check your question. As it is written, the first statement does not lead to the second. Here are the steps. For clarity, I'm going to say (F/B)^(μ*cotα) = Z, because it's a big messy term that appears in both statements. It won't get broken up, just moved around; it doesn't deserve more than one letter.
(F/B)^(μ*cotα) = (2μ*P*cotα + 2Y)/Y
Z = (2μ*P*cotα + 2Y)/Y
---------------------------------multiply by Y
YZ = 2μ*P*cotα + 2Y
---------------------------------subtract 2Y
YZ-2Y = 2μ*P*cotα
---------------------------------divide by 2μ and factor out Y
Y(Z-2)/2μ = P*cotα
---------------------------------cotα = 1/tanα, so multiply by tanα
Y(tanα)(Z-2)/2μ = P
---------------------------------now P is separated, as in the answer, and so is Y.
I'm going to take a sideways step now, to show why I don't think this is soluble. In a standard proof, I'd keep manipulating this P to look like the P I'm proving. Instead, I'm going to set them equal to each other, allowing me to cancel some stuff out and boil the problem down to its simplest elements. If it really is provable, then P always equals P and these two P equivalents should cancel out to 1 = 1.
P = Y(tanα)(Z-2)/2μ = Y(1+((tanα)/μ))(1-Z) ------ remember the Z substitution I made in the first step.
---------------------------------divide by Y
(tanα)(Z-2)/2μ = (1+((tanα)/μ))(1-Z)
---------------------------------there aren't any binomials the same between right and left (Z-2 and Z-1 are very different animals), but the right side binomials can be multiplied together.
(tanα)(Z-2)/2μ = 1 + (tanα)/μ - Z - Z((tanα)/μ)
Here it looks pretty nasty. Two terms added together on the left, four terms added together on the right, that's almost impossible to cancel out. But maybe expanding Z will help us. I'm going back a step to before the binomial expansion.
(tanα)([(F/B)^(μ*cotα)]-2)/2μ = (1+((tanα)/μ))(1-[(F/B)^(μ*cotα)])
---------------------------------simplify cot to 1/tan and space out some operations for clarity
([(F/B)^(μ/tanα)]-2) * (tanα)/2μ = (1+((tanα)/μ)) * (1-[(F/B)^(μ/tanα)])
The trouble here is, even though Z has α and μ in it, they're in an exponent. They're not coming out of there. We can't even arrange the equation around far enough to try to simplify it with logarithms. We might as well keep calling it Z.
I can't move the problem past this point. In my opinion it's either it's miswritten or it's insoluble.
Edit: Dang, while I was working all this out, Michael went and worked out what the typo was. Curse him for competence!
(F/B)^(μ*cotα) = (2μ*P*cotα + 2Y)/Y multiply by Y/2 to get:
(Y/2)(F/B)^(μ*cotα) = μ*P*cotα + Y, subtract Y to get:
(Y/2)(F/B)^(μ*cotα) -Y = μ*P*cotα, factor out Y to get:
Y[ (1/2)(F/B)^(μ*cotα) -1 ] = μ*P*cotα, divide by μ*cotα to get:
Y[ (1/2)(F/B)^(μ*cotα) -1 ] * 1/(μ*cotα) = P, rewrite 1/(μ*cotα) as tanα /μ
p=Y(tanα/μ)*[ (1/2)(F/B)^(μ*cotα) -1 ].
Not quite what you asked for but this is correct.
(what you asked for is not correct)
hi chum this an uncomplicated question in line with linear equations you ought to use transposition approach in this technique while a term is transfered from one section to different its sign turns into opposite answer: x+4-7x=22 =4-7x=22-x (right here x comes from the left section its sign adjustments) =4=22-x+7x(right here 7x comes from the left section its sign adjustments) =4-22=6x(same) =-18=6x x=-18/6 =-3 Bye Bye!!!!!!!!!!!!!!!!!!!!!!