Hard thinking questions

zgozvrm

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

94JZA80

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.

tiny-tim

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.

zgozvrm

My bad. I both misread your equations and miscalculated mine.

You are absolutely right!

I Like Pi

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

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

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

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

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

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

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
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

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

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

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

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

Have a goodnight tiny-tim!

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

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



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

it makes sense!, thanks

And don't worry about it! you've helped me enough enjoy your dinner! Take care!