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

[94JZA80]

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]

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!

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! 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]

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

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]

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]

Hi I Like Pi!

(just got up :zzz: …)

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]

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 infinitely 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?

 

[tiny-tim]

Hi 94JZA80!

(have a pi: π )

sinθ = sin(π - θ)

(so what's the equivalent for cos?)

 

[94JZA80]

Hi 94JZA80!

(have a pi: π )

sinθ = sin(π - θ)

(so what's the equivalent for cos?)

right...
- if A = pi/6, then B must = 5pi/6 in order to confidently say that sin A = sin B.
- if A = pi/4, then B must = 3pi/4 in order to confidently say that sin A = sin B.
- if A = pi/3, then B must = 2pi/3 in order to confidently say that sin A = sin B.
- and so in general, if A = θ, then B must = (pi-θ) in order to confidently say that sin A = sin B.

likewise...
- if A = pi/6, then B must = 11pi/6 in order to confidently say that cos A = cos B.
- if A = pi/4, then B must = 7pi/4 in order to confidently say that cos A = cos B.
- if A = pi/3, then B must = 5pi/3 in order to confidently say that cos A = cos B.
- and so in general, if A = θ, then B must = (2pi-θ) in order to confidently say that cos A = cos B.

in other words, in order to show the symmetry of sin θ about pi/2 and 3pi/2 (or symmetry about the y-axis on the unit circle), one has to show that sin θ = sin (pi-θ). and in order to show the symmetry of cos θ about pi (or symmetry about the x-axis on the unit circle), one has to show that cos θ = cos (2pi-θ)...although from what I've found, most sources cite the supplemental symmetry cos identity alternatively as -(cos θ) = cos (pi - θ). i suppose i could have just looked up the identities, but its always good to understand the derivations...at any rate, I'm still have a difficult time seeing how this is getting us closer to solving for b.

 

[tiny-tim]

sorry!

sorry I've taken so long to reply, i seem to have missed your answer

I think we started with cos(2b-π/8) = sin(6b+π/8),
which you can rewrite as cos(2b-π/8) = cos(π/2 -6b-π/8) = cos(3π/8 -6b);

now you know that cosA = cosB if B = 2nπ ± A;

so 2b - π/8 = 2nπ ± (3π/8 - 6b) …

carry on from there ​

 

[94JZA80]

ahh, i see...i forgot to bring the integer variable n into the picture, and i failed to expound on the idea that cos A = cos B at regular intervals, despite pondering it heavily in my last 2 posts lol...

so if cos A = cos B, then B must = 2npi-A -->
B = 2npi-A
2b-pi/8 = 2npi-(3pi/8-6b)
2b-pi/8 = 2npi-3pi/8+6b
-4b = 2npi-pi/4
b = pi/16-npi/2

thanks for the insight...thinking in terms of the integer variable n is the only way to calculate a true general solution to b over the interval (-∞,∞). otherwise, you're just solving for specific solutions to b on the interval 0-2pi.

 

[tiny-tim]

so if cos A = cos B, then B must = 2npi-A -->
B = 2npi-A
2b-pi/8 = 2npi-(3pi/8-6b)
2b-pi/8 = 2npi-3pi/8+6b
-4b = 2npi-pi/4
b = pi/16-npi/2

(what happened to that π i gave you? )

yes, but don't forget you have to solve it for both versions of the ±

g'night! :zzz:​

 

[94JZA80]

(what happened to that π i gave you? )

yes, but don't forget you have to solve it for both versions of the ±

g'night! :zzz:​

LOL i was too lazy to copy-n-paste it b/c i was using that command to copy-n-paste a much larger expression at that moment...although i'll have you know that your sig has come in handy in the brief time I've been here anyways, here is the work showing that the first value i calculated, b = pi/16-npi/2, is correct:
cos(2b-pi/8) = sin(6b+pi/8)
cos[2(pi/16-npi/2)-pi/8] = sin[6(pi/16-npi/2)+pi/8]
cos(pi/8-npi-pi/8) = sin(3pi/8-3npi+pi/8)
cos(-npi) = sin(pi/2-3npi)

--> to show that the above general expression is true, we test it by substituting various integers for n:
n = 0: cos 0 = 1 & sin(pi/2) = 1
n = 1: cos(-pi) = -1 & sin(pi/2-3pi) = sin(-5pi/2) = -1
n = 2: cos(-2pi) = 1 & sin (pi/2-6pi) = sin(-11pi/2) = 1
...and here is the work showing that the other value i calculated, b = npi/4+pi/16, is correct:
B = 2npi+A
2b-pi/8 = 2npi+(3pi/8-6b)
2b-pi/8 = 2npi+3pi/8-6b
8b = 2npi+pi/2
b = npi/4+pi/16

...now, by letting b = npi/4+pi/16 and substituting again, we have:
cos(2b-pi/8) = sin(6b+pi/8)
cos[2(npi/4+pi/16)-pi/8] = sin[6(npi/4+pi/16)+pi/8]
cos(npi/2+pi/8-pi/8) = sin(3npi/2+3pi/8+pi/8)
cos(npi/2) = sin(3npi/2+pi/2)

--> and again, to show that the above general expression is true, we test it by substituting various integers for n:
n = 0: cos 0 = 1 & sin(pi/2) = 1
n = 1: cos(pi/2) = 0 & sin(3pi/2+pi/2) = sin(2pi) = 0
n = 2: cos(pi) = -1 & sin (6pi/2+pi/2) = sin(7pi/2) = -1...and so we have shown that there are two distinct values for b (pi/16-npi/2 and npi/4+pi/16) that create angles A & B such that cos A = sin B.

 

[tiny-tim]

...now I'm not entirely sure why we have to solve for b in both equations (B = 2npi-A and B = 2npi+A), especially when finding value(s) for b such that B = 2npi-A is enough to show that cos A = cos B...granted, i understand that showing that cos B = cos (2pi+A) is proof of the "2pi periocity/shift" identity. nevertheless, i went ahead and solved the alternate equation B = 2npi+A for b, and came up with the following:
B = 2npi+A
2b-pi/8 = 2npi+(3pi/8-6b)
2b-pi/8 = 2npi+3pi/8-6b
8b = 2npi+pi/4
b = npi/4+pi/32

Hi Eric!

You need both versions of the ± because the question asks you for all the solutions!

(and your second solution didn't check out because you subtracted the π/8 instead of adding it, going from your line 3 to 4 )

 

[94JZA80]

Hi Eric!

You need both versions of the ± because the question asks you for all the solutions!

(and your second solution didn't check out because you subtracted the π/8 instead of adding it, going from your line 3 to 4 )

i knew it! its always something stupid like that...i stared at my work for 15 minutes and didn't see the arithmetic error . i'll go back and edit my previous post to show that b = npi/4+pi/32 is also a workable solution.

thanks again,
Eric