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#37
Oct2210, 03:45 AM

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Hi 94JZA80!
(have a pi: π ) sinθ = sin(π  θ) (so what's the equivalent for cos?) 


#38
Oct2210, 10:02 AM

P: 119

 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 = (piθ) 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 = (2piθ) 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 yaxis on the unit circle), one has to show that sin θ = sin (piθ). and in order to show the symmetry of cos θ about pi (or symmetry about the xaxis on the unit circle), one has to show that cos θ = cos (2piθ)...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. 


#39
Oct2810, 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(2bπ/8) = sin(6b+π/8), which you can rewrite as cos(2bπ/8) = cos(π/2 6bπ/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 


#40
Oct2810, 06:00 PM

P: 119

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 = 2npiA > B = 2npiA 2bpi/8 = 2npi(3pi/86b) 2bpi/8 = 2npi3pi/8+6b 4b = 2npipi/4 b = pi/16npi/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 02pi. 


#41
Oct2810, 06:19 PM

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


#42
Oct2810, 11:43 PM

P: 119

anyways, here is the work showing that the first value i calculated, b = pi/16npi/2, is correct: cos(2bpi/8) = sin(6b+pi/8) cos[2(pi/16npi/2)pi/8] = sin[6(pi/16npi/2)+pi/8] cos(pi/8npipi/8) = sin(3pi/83npi+pi/8) cos(npi) = sin(pi/23npi) > 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/23pi) = sin(5pi/2) = 1 n = 2: cos(2pi) = 1 & sin (pi/26pi) = 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 2bpi/8 = 2npi+(3pi/86b) 2bpi/8 = 2npi+3pi/86b 8b = 2npi+pi/2 b = npi/4+pi/16 ...now, by letting b = npi/4+pi/16 and substituting again, we have: cos(2bpi/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/8pi/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/16npi/2 and npi/4+pi/16) that create angles A & B such that cos A = sin B. 


#43
Oct2910, 06:58 AM

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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 ) 


#44
Oct2910, 09:45 AM

P: 119

thanks again, Eric 


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