Sine and Cosine Addition FormulasDate: 11/29/2000 at 00:55:14 From: Phara Subject: Deriving addition formulas for trig How do you prove this? sin(x+y) = sin(x)cos(y) + cos(x)sin(y) What about cos(x+y)? If sin(2Q) = a, then what is sin(Q) + cos(Q) in terms of a? Date: 11/29/2000 at 12:14:08 From: Doctor Rob Subject: Re: Deriving addition formulas for trig Thanks for writing to Ask Dr. Math, Phara. To derive the above identity, draw a diagram as follows. Start with a Cartesian rectangular coordinate system with origin O. Pick a point A on the positive x-axis. Construct ray OB so that <AOB has measure x. Construct ray OC so that <BOC has measure y. Then <AOC has measure x + y. Erect a perpendicular to OA at A intersecting OB at D. Erect a perpendicular to OB at D intersecting OC at E. Drop a perpendicular from E to OA, intersecting it at F, and intersecting OD at G. Now you have several right triangles, OAD, ODE, OFE, OFG, and GDE, each containing one of the angles with measure x, y, and x + y: ^ /C | E/ | /\ | /|x\ | / | \ | / | \ | / | \ B | / | \ _,-' | / | \D _,-' | / | _,o' | | / G| _,-' | | / _,+' | | /y _,-' | | |/_,-'x | | --o-----------o-------o---> O| F A Now let OE have length r. Then, using triangle ODE and the definitions of sine and cosine, OD = r*cos(y) DE = r*sin(y) Next, using triangle GDE, EG = r*sin(y)/cos(x) DG = r*sin(y)*tan(x) That implies that OG = OD - DG = r*cos(y) - r*sin(y)*tan(x) = r*[cos(y)-sin(y)*tan(x)] Now using triangle OFG, FG = OG*sin(x) = r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)] That tells you the length EF = EG + FG = r*sin(y)/cos(x) + r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)] = r*[sin(x)*cos(y)-sin(y)*sin^2(x)/cos(x)+sin(y)/cos(x)] = r*[sin(x)*cos(y)+sin(y)*[1-sin^2(x)]/cos(x)] = r*[sin(x)*cos(y)+sin(y)*cos^2(x)/cos(x)] = r*[sin(x)*cos(y)+cos(x)*sin(y)] On the other hand, using triangle OFE, EF = r*sin(x+y) Setting these last two equal and dividing out the r gives sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y) To derive the cosine identity, consider cos(x+y) = sin(pi/2-[x+y]) = sin([pi/2-x]+[-y]) Now use the sine identity proved above: cos(x+y) = sin(Pi/2-x)*cos(-y) + cos(Pi/2-x)*sin(-y) = cos(x)*cos(y) - sin(x)*sin(y) For your other question, sin(2*Q) = a 2*sin(Q)*cos(Q) = a 1 + 2*sin(Q)*cos(Q) = 1 + a sin^2(Q) + 2*sin(Q)*cos(Q) + cos^2(Q) = 1 + a [sin(Q)+cos(Q)]^2 = 1 + a sin(Q) + cos(Q) = +-sqrt(1+a) The sign to use is determined by which quadrant Q + pi/4 might lie in. If Q + pi/4 is in the first or second quadrant, then the positive sign is correct, and otherwise the negative sign is the one to use. This is because sin(Q) + cos(Q) = sqrt(2)*sin(Q+Pi/4) - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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