proof left as an exercise
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07-01-2022, 07:51 PM
(This post was last modified: 07-06-2022 01:33 PM by Albert Chan.)
Post: #10
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RE: proof left as an exercise
Code: sin((2*k-1)*x) / sin(x) = 1 + 2*[cos(2x) + cos(4x) + ... + cos((2*k-2)*x)] Another way, using above sin(n*x)/sin(x) identities 1 + 4*cos(20°) = 2*cos(20°) + 2*(cos(60°) + cos(20°)) = 2*cos(20°) + 4*cos(20°)*cos(40°) = 2*cos(20°) * (2*cos(40°)+1) = 2*cos(20°) * sin(60°)/sin(20°) = 2*sin(60°) / tan(20°) --> 2*cos(30°) / (1 + 4*sin(70°)) = tan(20°) --- Using complementary angle, we can get ratio of cos over cos (or, sin over cos) XCAS> Q := makelist(k->f(k*x),1,4) XCAS> normal(Q(f=sin, x=pi/2-y)) → [cos(y),sin(2*y),-cos(3*y),-sin(4*y)] XCAS> normal(Q(f=cos, x=pi/2-y)) → [sin(y),-cos(2*y),-sin(3*y),cos(4*y)] (x = pi/2-y) transform affected both side of identity; it is better to reverse sum order. Code: cos((2k-1)*y) / cos(y) = 2*[cos((2k-2)*y) - cos((2k-4)*y) + cos((2k-6)*y) - ... ] + (-1)^(k+1) Example, redo the proof, using cos over cos identity: cos(3x)/cos(x) = (2*cos(2x) - 1) 1 + 4*cos(20°) = 2*cos(60°) + 4*cos(20°) = 2*cos(20°)*(2*cos(40°)-1) + 4*cos(20°) = 2*cos(20°) * (2*cos(40°)+1) = 2*cos(20°) * sin(60°)/sin(20°) = 2*sin(60°) / tan(20°) |
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Messages In This Thread |
proof left as an exercise - Thomas Klemm - 06-06-2022, 11:41 PM
RE: proof left as an exercise - Ángel Martin - 06-07-2022, 05:05 AM
RE: proof left as an exercise - Thomas Klemm - 06-07-2022, 05:32 AM
RE: proof left as an exercise - Albert Chan - 06-07-2022, 05:36 PM
RE: proof left as an exercise - Albert Chan - 06-07-2022, 06:17 PM
RE: proof left as an exercise - Albert Chan - 06-08-2022, 01:50 AM
RE: proof left as an exercise - Albert Chan - 06-08-2022, 11:12 AM
RE: proof left as an exercise - Thomas Klemm - 06-08-2022, 11:18 PM
RE: proof left as an exercise - Albert Chan - 06-09-2022, 12:35 AM
RE: proof left as an exercise - Albert Chan - 07-01-2022 07:51 PM
RE: proof left as an exercise - Albert Chan - 07-02-2022, 11:44 PM
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