Hard thinking questions


by I Like Pi
Tags: advanced functions, trigonometry
tiny-tim
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#37
Oct22-10, 03:45 AM
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Hi 94JZA80!

(have a pi: π )

sinθ = sin(π - θ)

(so what's the equivalent for cos?)
94JZA80
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Oct22-10, 10:02 AM
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Quote Quote by tiny-tim View Post
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
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Oct28-10, 11:05 AM
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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
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Oct28-10, 06:00 PM
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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
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#41
Oct28-10, 06:19 PM
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Quote Quote by 94JZA80 View Post
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!
94JZA80
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#42
Oct28-10, 11:43 PM
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Quote Quote by tiny-tim View Post
(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!
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
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Oct29-10, 06:58 AM
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Quote Quote by 94JZA80 View Post
...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
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#44
Oct29-10, 09:45 AM
P: 114
Quote Quote by tiny-tim View Post
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


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