trigonometrik baÄ�Ä±ntÄ±lar

# trigonometrik baÄ�Ä±ntÄ±lar - cos = cos(- ) = sin(...

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cos θ = cos(- θ ) = sin( π /2 - θ ) = cosh j θ = ½(e j θ + e -j θ ) sin θ = - sin(- θ ) = cos( π /2 - θ ) = - j sinh θ = (e j θ - e -j θ )/2j tan θ = - tan(- θ ) = cot( π /2 - θ ) = -j tanh j θ = ½(e j θ - e -j θ )/ j(e j θ + e -j θ ) For small θ :- sin θ = θ - θ 3 /3! + θ 5 /5! - . .. cos θ = 1 - θ 2 /2! + θ 4 /4! - . .. tan θ = θ + θ 3 /3 + 2 θ 5 /15 + . .. atan θ = θ - θ 3 /3 + θ 5 /5 + . .. sin 2 θ = 2 cos θ sin θ cos 2 θ = cos 2 θ - sin 2 θ = 2 cos 2 θ - 1 = 1 - 2 sin 2 θ tan 2 θ = 2 tan θ / (1 - tan 2 θ ) sin θ /2 = ± [(1 - cos θ )/2] cos θ /2 = ± [(1 + cos θ )/2] tan θ /2 = sin θ /(1 + cos θ ) sin 3 θ = 3 sin θ - 4 sin 3 θ cos 3 θ = - 3 cos θ + 4 cos 3 θ tan 3 θ = (3 tan θ - tan 3 θ )/ (1 - 3 tan 2 θ ) cos 2 θ + sin 2 θ = 1 sec 2 θ - tan 2 θ = 1 cosec 2 θ - cot 2 θ = 1 sin 2 θ = ½(1 - cos 2 θ ) cos 2 θ = ½(1 + cos 2 θ ) tan 2 θ = (1 - cos 2 θ ) /(1 + cos 2 θ ) sin(A + B) = sin A cos B + cos A sin B sin(A - B) = sin A cos B - cos A sin B

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## This note was uploaded on 01/03/2010 for the course MATH 123 taught by Professor Sait during the Spring '09 term at Istanbul Technical University.

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trigonometrik baÄ�Ä±ntÄ±lar - cos = cos(- ) = sin(...

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