
#19
Oct2110, 06:04 PM

P: 113





#20
Oct2110, 06:10 PM

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P: 26,167

Hi I Like Pi!!
(since you like them so much, have a pi: π ) 



#21
Oct2110, 06:12 PM

P: 754

My bad. I both misread your equations and miscalculated mine.
You are absolutely right! 



#22
Oct2110, 06:15 PM

P: 91

well i tried that.... i used cos instead: [tex] cos(2bpi/8) = cos[pi/2(6b+pi/8)] [/tex] [tex] cos(2bpi/8)  cos[pi/2(6b+pi/8)] = 0 [/tex] [tex] cos[(2bpi/8)(pi/26bpi/8)] = 0 [/tex] [tex] 8bpi/2 = cosinverse(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)... 



#23
Oct2110, 06:15 PM

P: 113

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 2a1, AND when 2a1 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
Oct2110, 06:18 PM

P: 91





#25
Oct2110, 06:23 PM

P: 754

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
Oct2110, 06:24 PM

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P: 26,167

You're saying cos(A)  cos(B) = 0 if cos(AB) = 0, which isn't correct.
Start again draw a graph and look at it when does cos(A) = cos(B)? 



#27
Oct2110, 06:29 PM

P: 113

sin(6b+pi/8) = cos(2bpi/8) then, using the identity cosθ = sin(π/2  θ), you have sin(6b+pi/8) = sin(pi/2(2bpi/8)). therefore, 6b+pi/8 = pi/22b+pi/8. now solve for b. 



#28
Oct2110, 06:30 PM

P: 91

then you bring that over? So, cos(A)cos(B) = 0 (am I right here?) and then you group similar terms, so cos(AB) = 0? why isn't it correct? Thanks tim i really appreciate your time! 



#29
Oct2110, 06:36 PM

P: 91

again, i thank you dearly! 



#30
Oct2110, 06:39 PM

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P: 26,167

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 ) 



#31
Oct2110, 06:40 PM

P: 113

*EDIT*  it appears tinytim 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. 



#32
Oct2110, 07:04 PM

P: 91

Have a goodnight tinytim! 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 ! 



#33
Oct2110, 07:15 PM

P: 113

so go back to this equation: sin(6b+pi/8) = cos(2bpi/8) ...and through manipulation: sin(6b+pi/8) = sin(pi/2(2bpi/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... 



#34
Oct2110, 07:32 PM

P: 91





#35
Oct2210, 01:47 AM

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P: 26,167

Hi I Like Pi!
(just got up ) 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; 



#36
Oct2210, 03:08 AM

P: 113

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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