Mastering Hard Thinking Questions: Solving for a, p, and q with Given Equations

In summary, the given equations can be simplified to csc(6b + π/8) = sec(2b - π/8). To solve for b, we can use the identities cscx = 1/sinx and secx = 1/cosx. By cross-multiplying, we get cos(2b-π/8) = sin(6b+π/8). From this, we can determine that 2b-π/8 = 6b+π/8. Solving for b, we get b = π/10.
  • #1
I Like Pi
91
0

Homework Statement



a) given, a + 5 and 2a-1, one is 40% greater than the other one, solve for a.

b) given, 5x[tex]^{4}[/tex] + 4x[tex]^{3}[/tex] + 3x[tex]^{2}[/tex] + Px + Q. When divided by x[tex]^{2}[/tex] - 1, you get a remainder of 0, solve for p and q.

c) given csc(6b + [tex]\frac{pi}{8}[/tex]) = sec(2b - [tex]\frac{pi}{8}[/tex]), solve for b.

Homework Equations

The Attempt at a Solution



please help, i can't get started on these three at all... :confused:

edit:

c) i know that cscx = 1/sinx and secx = 1/cosx, i then cross multiplied to get [tex]cos(2b-pi/8) = sin(6b+pi/8)...
 
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  • #2
Question a) shouldn't be a problem at all.
Any attempt by you ?
 
  • #3
Quinzio said:
Question a) shouldn't be a problem at all.
Any attempt by you ?

Would it be (a+5) = .4(2a-1) + 2a-1 ? or .4(a+5) + (a+5) = 2a - 1 ?
 
  • #4
I Like Pi said:
Would it be (a+5) = .4(2a-1) + 2a-1 ? or .4(a+5) + (a+5) = 2a - 1 ?

It's not clear from the text. Choose one and go on, or solve both in sequence.
 
  • #5
Quinzio said:
It's not clear from the text. Choose one and go on, or solve both in sequence.

well then you get a+5 = 0.8a - 0.4 + 2a -1
a+5 = 2.8a - 1.4
1.8a - 6.4 = 0
1.8a = 6.4
a = 3.55555556 :confused:

or

2a - 1 = .4a + 2 + a + 5
2a - 1 = 1.4a + 7
0.6a - 8 = 0
0.6a = 8
a = 13.3333333

therefore a can be 13.333 or 3.555 ? :confused:
 
  • #6
yes.
 
  • #7
Quinzio said:
yes.

thanks :smile: could you help me get started on b) ?

Thanks for your time!
 
  • #8
b)
Make this multiplication [tex] (x^2-1)(Ax^2+Bx+C)[/tex]
 
  • #9
Quinzio said:
b)
Make this multiplication [tex] (x^2-1)(Ax^2+Bx+C)[/tex]

that would give you [tex] ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]. Is that what you mean?
 
  • #10
I Like Pi said:
that would give you [tex] ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]. Is that what you mean?

Ok, so now give the solution. It's really easy.
 
  • #11
Quinzio said:
Ok, so now give the solution. It's really easy.

do you use the original equation? so a = 5, b = 4, and c = 3?

but that doesn't work... cause then you get:[tex]5x^4 + 4x^3 + (3 - 5)x^2 - 4x - 3[/tex]
 
  • #12
You realize by yourself there's something wrong. Ok.

You should come alone to the solution. The next step would be giving you the solution.
 
  • #13
Quinzio said:
You realize by yourself there's something wrong. Ok.

You should come alone to the solution. The next step would be giving you the solution.

I don't see what I could possibly do to get the answer :confused: I know that c - a = 3, that's all... but c itself is 3 :confused:
 
  • #14
Why you say "c itself is 3"... ?
Tell me where is written that ?
 
  • #15
Quinzio said:
Why you say "c itself is 3"... ?
Tell me where is written that ?

oh, well i based it on the original... a = 5, b = 4, c = 3...

well, then if that's the case,

[tex]ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]
[tex]=5x^4 + 4x^3 + (c - 5)x^2 - 4x - c[/tex]
therefore
[tex]
c = 8 [/tex]?
and therefore:
[tex]P = -4[/tex]
[tex]Q = -8[/tex]

is that what you meant?

Thanks for your help! Means a lot!
 
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  • #16
I Like Pi said:
well then you get a+5 = 0.8a - 0.4 + 2a -1
a+5 = 2.8a - 1.4
1.8a - 6.4 = 0
1.8a = 6.4
a = 3.55555556 :confused:

or

2a - 1 = .4a + 2 + a + 5
2a - 1 = 1.4a + 7
0.6a - 8 = 0
0.6a = 8
a = 13.3333333

therefore a can be 13.333 or 3.555 ? :confused:

Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?
 
  • #17
zgozvrm said:
Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?

Well you would do .4 * 20 + 20, a value 40% greater then 20 is 28?
 
  • #18
Yes, but that simplifies...

0.4 * 20 + 20 = (0.4 + 1) * 20 = 1.4 * 20

Therefore, a number that is 40% larger than x is 1.4 * x
 
  • #19
zgozvrm said:
Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?

well at first glance (and i do mean i only "glanced" at it for a split second), i thought the same thing...here are the equations he set up:
I Like Pi said:
(a+5) = .4(2a-1) + 2a-1
.4(a+5) + (a+5) = 2a - 1

look closer at the right-hand side of the first equation. 0.4(2a-1) + 2a-1 = 1.4(2a-1), and the original equation becomes a+5 = 1.4(2a-1), which certainly implies that a+5 is 40% larger than 2a-1. likewise, we can see that the left-hand side of the 2nd equation, 0.4(a+5) + (a+5), equals 1.4(a+5). and so equation 2 becomes 1.4(a+5) = 2a-1, which certainly implies that 2a-1 is 40% greater than a+5.


I Like Pi said:
Well you would do .4 * 20 + 20, a value 40% greater then 20 is 28?

your calculations are correct - you just chose to represent the quantity on one side ofthe equation as 0.4x + x instead of 1.4x.
 
  • #20
Hi I Like Pi! :smile:

(since you like them so much, have a pi: π :wink:)
I Like Pi said:
c) given csc(6b + [tex]\frac{pi}{8}[/tex]) = sec(2b - [tex]\frac{pi}{8}[/tex]), solve for b.

c) i know that cscx = 1/sinx and secx = 1/cosx, i then cross multiplied to get [tex]cos(2b-pi/8) = sin(6b+pi/8)...

Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB. :smile:
 
  • #21
My bad. I both misread your equations and miscalculated mine.

You are absolutely right!
 
  • #22
tiny-tim said:
Hi I Like Pi! :smile:

(since you like them so much, have a pi: π :wink:)Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB. :smile:

haha, thank you :smile: !

well i tried that... i used cos instead:
[tex]
cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]
[/tex]
[tex]
cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0
[/tex]
[tex]
cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0
[/tex]
[tex]
8b-pi/2 = cos-inverse(0)
[/tex]
[tex]
8b = pi
[/tex]
[tex]
b = pi/8
[/tex]

it doesn't work :frown:

edit: I used geometer's sketchpad to graph the two, and the first point of intersection is (1.76715.., -1)...
 
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  • #23
what confuses me about the first question is that i am not so sure that a should have two distinct values. sure, if the text (or wherever the OP got the problem from) literally does not give any more information than what he posted (i.e. doesn't tell you which quantity is 40% greater than the other), then i can see how you would have to solve for a in two different situations, yielding two disticnt values for a (that is, when a+5 is 40% greater than 2a-1, AND when 2a-1 is 40% greater than a+5). either way, the OP has clearly shown us that he knows how to solve for a variable using arithmetic and algebraic manipulations, and has been able to determine a value for a regardless of which quantity is 40% larger than the other.
 
  • #24
94JZA80 said:
well at first glance (and i do mean i only "glanced" at it for a split second), i thought the same thing...here are the equations he set up:


look closer at the right-hand side of the first equation. 0.4(2a-1) + 2a-1 = 1.4(2a-1), and the original equation becomes a+5 = 1.4(2a-1), which certainly implies that a+5 is 40% larger than 2a-1. likewise, we can see that the left-hand side of the 2nd equation, 0.4(a+5) + (a+5), equals 1.4(a+5). and so equation 2 becomes 1.4(a+5) = 2a-1, which certainly implies that 2a-1 is 40% greater than a+5.




your calculations are correct - you just chose to represent the quantity on one side ofthe equation as 0.4x + x instead of 1.4x.

94JZA80 said:
what confuses me about the first question is that i am not so sure that a should have two distinct values. sure, if the text (or wherever the OP got the problem from) literally does not give any more information than what he posted (i.e. doesn't tell you which quantity is 40% greater than the other), then i can see how you would have to solve for a in two different situations, yielding two disticnt values for a (that is, when a+5 is 40% greater than 2a-1, AND when 2a-1 is 40% greater than a+5). either way, the OP has clearly shown us that he knows how to solve for a variable using arithmetic and algebraic manipulations, and has been able to determine a value for a regardless of which quantity is 40% larger than the other.

Thanks very much for your help! and yes, I guess it's just a question that could go both ways because it is written like that. Though I was in a rush to write it down, i hope i didn't write it wrong, but either way, it is pretty straight forward now that i think of it.. :redface:
 
  • #25
I Like Pi said:
oh, well i based it on the original... a = 5, b = 4, c = 3...

well, then if that's the case,

[tex]ax^4 + bx^3 + (c - a)x^2 - bx - c[/tex]
[tex]=5x^4 + 4x^3 + (c - 5)x^2 - 4x - c[/tex]
therefore
[tex]
c = 8 [/tex]?
and therefore:
[tex]P = -4[/tex]
[tex]Q = -8[/tex]

is that what you meant?

Thanks for your help! Means a lot!

That's correct!

You can also attack this problem by doing the long division:

The dividend is [itex]5x^4 + 4x^3 + 3x^2 + Px + Q[/tex]
The divisor is [itex]x^2 - 1[/tex]

[itex]x^2[/tex] goes into [itex]5x^4[/tex]
[itex]5x^2[/tex] times.
(That is, [itex]x^2 \times 5x^2 = 5x^4[/tex])

So multiply the divisor by [itex]5x^2[/tex] and you get [itex](5x^4 - 5x^2)[/tex]
Subtract this from the dividend and you get [itex]4x^3 + 8x^2 + Px + Q[/tex]

Now, determine how many times [itex]x^2[/tex] goes into [itex]4x^3[/tex]: that would be [itex]4x[/tex] times.
Multiply [itex]4x \times (x^2 - 1) = (4x^3 - 4x)[/tex]

Subtract this from [itex]4x^3 + 8x^2 + Px + Q[/tex] and you get [itex]8x^2 + (P + 4)x + Q[/tex]

Continuing...
[itex]8 \times (x^2 - 1) = (8x^2 - 8)[/tex]
[itex]8x^2 + (P + 4)x + Q - (8x^2 - 8) = (P + 4)x + (Q + 8)[/tex]

Since there is no remainder, (P + 4) and (Q + 8) must both be equal to 0.


Quinzio's way is MUCH easier, I just wanted to offer an alternative way.
 
  • #26
You're saying cos(A) - cos(B) = 0 if cos(A-B) = 0, which isn't correct.

Start again … draw a graph and look at it …

when does cos(A) = cos(B)?​
 
  • #27
I Like Pi said:
haha, thank you :smile: !

well i tried that... i used cos instead:
[tex]
cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]
[/tex]
[tex]
cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0
[/tex]
[tex]
cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0
[/tex]
[tex]
8b-pi/2 = cos-inverse(0)
[/tex]
[tex]
8b = pi
[/tex]
[tex]
b = pi/8
[/tex]

it doesn't work :frown:

i think you're trying to manipulate the equation more than you have to. leave it in the following form:

sin(6b+pi/8) = cos(2b-pi/8)


then, using the identity cosθ = sin(π/2 - θ), you have sin(6b+pi/8) = sin(pi/2-(2b-pi/8)). therefore, 6b+pi/8 = pi/2-2b+pi/8. now solve for b.
 
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  • #28
zgozvrm said:
That's correct!

You can also attack this problem by doing the long division:

The dividend is [itex]5x^4 + 4x^3 + 3x^2 + Px + Q[/tex]
The divisor is [itex]x^2 - 1[/tex]

[itex]x^2[/tex] goes into [itex]5x^4[/tex]
[itex]5x^2[/tex] times.
(That is, [itex]x^2 \times 5x^2 = 5x^4[/tex])

So multiply the divisor by [itex]5x^2[/tex] and you get [itex](5x^4 - 5x^2)[/tex]
Subtract this from the dividend and you get [itex]4x^3 + 8x^2 + Px + Q[/tex]

Now, determine how many times [itex]x^2[/tex] goes into [itex]4x^3[/tex]: that would be [itex]4x[/tex] times.
Multiply [itex]4x \times (x^2 - 1) = (4x^3 - 4x)[/tex]

Subtract this from [itex]4x^3 + 8x^2 + Px + Q[/tex] and you get [itex]8x^2 + (P + 4)x + Q[/tex]

Continuing...
[itex]8 \times (x^2 - 1) = (8x^2 - 8)[/tex]
[itex]8x^2 + (P + 4)x + Q - (8x^2 - 8) = (P + 4)x + (Q + 8)[/tex]

Since there is no remainder, (P + 4) and (Q + 8) must both be equal to 0.


Quinzio's way is MUCH easier, I just wanted to offer an alternative way.

Hey, thanks for the alternative! It makes much sense to me, especially with this, again, thank you for your time! :smile: its amazing the help i get here :smile:
tiny-tim said:
You're saying cos(A) - cos(B) = 0 if cos(A-B) = 0, which isn't correct.

Start again … draw a graph and look at it …

when does cos(A) = cos(B)?​

well, i got that cos(A) = cos(B) (i think that's right)
then you bring that over? So, cos(A)-cos(B) = 0 (am I right here?) and then you group similar terms, so cos(A-B) = 0? why isn't it correct?

Thanks tim :smile: i really appreciate your time!
 
  • #29
94JZA80 said:
i think you're trying to manipulate the equation more than you have to. leave it in the form sin(6b+pi/8) = cos(2b-pi/8). then, sin(6b+pi/8) = sin(pi/2+(2b-pi/8)). if the sine of those quantities are equal, then the quantities themselves are equal. that is, if sin A = sin B, then A = B. likewise, if cos A = cos B, then A = B. therefore, 6b+pi/8 = pi/2+2b-pi/8. now just solve for b.

hey! Thank you so much! It makes a lot of sense! would this be the case any time you have same trig = same trig, you would cancel it out? so if tan(x) = tan(x), x = x and you would solve? Don't you have to do the inverse to get rid of it :confused:

again, i thank you dearly!
 
  • #30
I Like Pi said:
well, i got that cos(A) = cos(B) (i think that's right)
then you bring that over? So, cos(A)-cos(B) = 0 (am I right here?) and then you group similar terms, so cos(A-B) = 0? why isn't it correct?

Well, for a start, if A = B, cos(A-B) = 1, isn't it?

Do what I said … draw a graph, and see when cosA = cosB !

(btw, it isn't only when A = B)

(and now I'm off to bed :zzz: …)
 
  • #31
I Like Pi said:
hey! Thank you so much! It makes a lot of sense! would this be the case any time you have same trig = same trig, you would cancel it out? so if tan(x) = tan(x), x = x and you would solve? Don't you have to do the inverse to get rid of it :confused:

again, i thank you dearly!

actually, now that you put it that way, i guess i can't definitively that if sin A = sin B, then A = B. for example, i can think of an instance in which sin A = sin B, but A != B...specifically, sin pi = sin 2pi = 0, but clearly pi != 2pi. in fact, sin pi = sin 2pi = sin 3pi = sin 4pi = sin npi, where n is any real integer, yet clearly each of these angles is unique and unequal to any other angle whose sine is also 0. perhaps tiny-tim could shed some light on this...i don't want to dole out false information if i can help it.



*EDIT* - it appears tiny-tim has shed some light on the subject, and he mentioned just what i was getting at above - that A does not necessarily equal B just b/c sin A = sin B. in fact, now that i think about it, sin & cos are operators just like addition and subraction, so you are correct that the inverse sin operation must be performed to both sides of an equation in order to rid one side of its sin operator.
 
Last edited:
  • #32
tiny-tim said:
Well, for a start, if A = B, cos(A-B) = 1, isn't it?

Do what I said … draw a graph, and see when cosA = cosB !

(btw, it isn't only when A = B)

(and now I'm off to bed :zzz: …)

Thank you! so, cos(A-B) = 1 is that by convention? or in this particular question?

Have a goodnight tiny-tim!

94JZA80 said:
actually, now that you put it that way, I'm not so sure that if sin A = sin B, then A = B. for example, i can think of an instance in which sin A = sin b, but A != B, specifically, sin pi = sin 2pi = 0, but clearly pi != 2pi. perhaps tiny-tim could shed some light on this...i don't want to dole out false information if i can help it.
*EDIT* - it appears tiny-tim has shed some light on the subject, and he mentioned just what i was getting at above - that A does not necessarily equal B just b/c sin A = sin B. perhaps it is best to do what he suggested and draw a diagram to see exactly when sin θ = cos θ. and you'll find that sin θ = sin θ for infinitely many values of A and B...and those values for A and B will be at regular intervals.

an extra hint: the diagram/graph you'll want to sketch is the "unit circle." you should find four different angles around the unit circle such that sin θ = con θ, one in each quadrant.

Thanks for the heads up! :smile: i guess i'll have to remember that cos(A)-cos(B) = 0 : cos(A - B) = 1 :confused:

Yes, i drew a graph, well used technology, but i want to see how to do it algebraically, without the need of a graph.. in case i don't have the time to draw one :/

Thanks so much :smile: !
 
  • #33
I Like Pi said:
so, cos(A-B) = 1 is that by convention? or in this particular question?

think about it...if A = B, then A-B = 0. hence cos(A-B) = cos 0 = 1



I Like Pi said:
Yes, i drew a graph, well used technology, but i want to see how to do it algebraically, without the need of a graph.. in case i don't have the time to draw one :/

Thanks so much :smile: !

well since i was initially incorrect in my assumption that if sin A = sin B, then A = B (since it is only sometimes true, and not always true), i have to assume that its necessary to perform the inverse trig operator in order to eliminate the trig operators on each side of the equation and solve for b algebraically. however, it appears to me that if you've been able to manipulate the equation into the form sin A = sin B, then applying the arcsin operator to both sides of the equation would yield A = B anyways. and so i think that i originally managed to get from one step to the next correctly, and was just lucky that my incorrect reasoning yielded the same result.

so go back to this equation:
sin(6b+pi/8) = cos(2b-pi/8)

...and through manipulation:
sin(6b+pi/8) = sin(pi/2-(2b-pi/8))

now that the equation is in the form sin A = sin B, you can apply your inverse sin (arcsin) operator to equate A and B, and thus solve for b. i don't want to go any further for fear of giving away the answer, or at least what i think could be the answer...
 
Last edited:
  • #34
94JZA80 said:
think about it...if A = B, then A-B = 0. hence cos(A-B) = cos 0 = 1

it makes sense!, thanks :smile:

i'll try and get to this later tonight...its dinner time!

And don't worry about it! you've helped me enough :smile: enjoy your dinner! Take care!
 
  • #35
Hi I Like Pi! :smile:

(just got up :zzz: …)
I Like Pi said:
Yes, i drew a graph, well used technology, but i want to see how to do it algebraically, without the need of a graph.. in case i don't have the time to draw one :/

uhhh? :redface: how long does it take to draw a wavy line? :confused:

If you draw it, you should see that any horizontal line cuts the cos graph twice in every 2π …

so if one of the cuts is at θ, the others will be at … ?

(and then try the same for the sin graph)

Do it this way (using the graph) first, so that you see what's going on;

once you've done that, try the alternative method, of the formula for cosA - cosB in the PF Library on trigonometric identities (94JZA80, you should know that too! :wink:).​
 
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