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Hard thinking questions

by I Like Pi
Tags: advanced functions, trigonometry
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94JZA80
#19
Oct21-10, 06:04 PM
P: 119
Quote Quote by zgozvrm View Post
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:
Quote Quote by I Like Pi View Post
(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.


Quote Quote by I Like Pi View Post
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.
tiny-tim
#20
Oct21-10, 06:10 PM
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Hi I Like Pi!!

(since you like them so much, have a pi: π )
Quote Quote by I Like Pi View Post
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.
zgozvrm
#21
Oct21-10, 06:12 PM
P: 754
My bad. I both misread your equations and miscalculated mine.

You are absolutely right!
I Like Pi
#22
Oct21-10, 06:15 PM
P: 91
Quote Quote by tiny-tim View Post
Hi I Like Pi!!

(since you like them so much, have a pi: π )


Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB.
haha, thank you !

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

edit: I used geometer's sketchpad to graph the two, and the first point of intersection is (1.76715.., -1)...
94JZA80
#23
Oct21-10, 06:15 PM
P: 119
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.
I Like Pi
#24
Oct21-10, 06:18 PM
P: 91
Quote Quote by 94JZA80 View Post
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.
Quote Quote by 94JZA80 View Post
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..
zgozvrm
#25
Oct21-10, 06:23 PM
P: 754
Quote Quote by I Like Pi View Post
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.
tiny-tim
#26
Oct21-10, 06:24 PM
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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)?
94JZA80
#27
Oct21-10, 06:29 PM
P: 119
Quote Quote by I Like Pi View Post
haha, thank you !

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
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.
I Like Pi
#28
Oct21-10, 06:30 PM
P: 91
Quote Quote by zgozvrm View Post
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! its amazing the help i get here
Quote Quote by tiny-tim View Post
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 i really appreciate your time!
I Like Pi
#29
Oct21-10, 06:36 PM
P: 91
Quote Quote by 94JZA80 View Post
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

again, i thank you dearly!
tiny-tim
#30
Oct21-10, 06:39 PM
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Quote Quote by I Like Pi View Post
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 …)
94JZA80
#31
Oct21-10, 06:40 PM
P: 119
Quote Quote by I Like Pi View Post
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

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.
I Like Pi
#32
Oct21-10, 07:04 PM
P: 91
Quote Quote by tiny-tim View Post
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 …)
Thank you! so, cos(A-B) = 1 is that by convention? or in this particular question?

Have a goodnight tiny-tim!

Quote Quote by 94JZA80 View Post
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 infintely 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! i guess i'll have to remember that cos(A)-cos(B) = 0 : cos(A - B) = 1

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 !
94JZA80
#33
Oct21-10, 07:15 PM
P: 119
Quote Quote by I Like Pi View Post
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



Quote Quote by I Like Pi View Post
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 !
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...
I Like Pi
#34
Oct21-10, 07:32 PM
P: 91
Quote Quote by 94JZA80 View Post
think about it...if A = B, then A-B = 0. hence cos(A-B) = cos 0 = 1
it makes sense!, thanks

i'll try and get to this later tonight...its dinner time!
And don't worry about it! you've helped me enough enjoy your dinner! Take care!
tiny-tim
#35
Oct22-10, 01:47 AM
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Hi I Like Pi!

(just got up …)
Quote Quote by I Like Pi View Post
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? how long does it take to draw a wavy line?

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! ).
94JZA80
#36
Oct22-10, 03:08 AM
P: 119
i'm currently experiencing a major mental blockage...when i graph either the sine function or the cosine function, i see that any horizontal line intersects either graph twice every 2pi radians...though i'm not sure what the significance of that is, especially considering that the interval between intersects varies as the value of the function varies. i did observe though that sin A = sin B every 2pi radians, and cos A = cos B every 2pi radians. that is, sin θ = sin (2npi+θ) where n is any real integer, and cos θ = cos (2npi+θ) where n is any real integer. so i'm beginning to question whether it even matters that the value of sin or cos is the same for an infintely many angles 2pi radians apart. if sin θ = sin (2pi+θ) = sin (4pi+θ) = sin (6pi+θ) = sin (2npi+θ), why can't we simply calculate a value for b choosing just one of the above angles?


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