#title 삼각함수의 정리 == 삼각함수의 정리 == * 작성자 조재혁([mailto:minzkn@minzkn.com]) * 고친과정 2008년 어느날 : 처음씀 === 덧셈정리 === {{{ sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y) sin(x - y) = sin(x) * cos(y) - cos(x) * sin(y) cos(x + y) = cos(x) * cos(y) - sin(x) * sin(y) cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y) tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y)) tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x) * tan(y)) }}} === 배각정리 === {{{ sin(2 * x) = 2 * sin(x) * cos(x) cos(2 * x) = cos²(x) - sin²(x) = 2 * cos²(x) - 1 = 1 - 2 * sin²(x) tan(2 * x) = (2 * tan(x)) / (1 - tan²(x)) }}} === 반각의 정리 === {{{ sin²(x / 2) = (1 - cos(x)) / 2 cos²(x / 2) = (1 + cos(x)) / 2 tan²(x / 2) = (1 - cos(x)) / (1 + cos(x)) }}} === 삼각함수 공식 총괄정리 === {{{ sin(-θ)=-sinθ, cos(-θ)=cosθ, tan(-θ)=-tanθ sin²θ+cos²θ=1, sec²θ-tan²θ=1, csc²θ-cot²θ=1 sin(π/2-θ)=cosθ, sin(π/2+θ)=cosθ, sin(θ±π/2)=±cosθ cos(π/2-θ)=sinθ, cos(π/2+θ)=-sinθ, cos(θ±π/2)=Ŧsinθ sin(π-θ)=sinθ, sin(π+θ)=-sinθ, sin(θ±π)=-sinθ cos(π-θ)=-cosθ, cos(π+θ)=-cosθ, cos(θ±π)=-cosθ sin(α±β)=sinαcosβ±cosαsinβ, cos(α±β)=cosαcosβŦsinαsinβ tan(α+β)=tanα+tanβ/1-tanαtanβ, tan(α-β)=tanα-tanβ/1+tanαtanβ sin2θ=2sinθcosθ cos2θ=cos²θ-sin²θ=2cos²θ-1=1-2sin²θ, tan2θ=2tanθ/1-tan²θ sin²θ/2=1-cosθ/2, cos²θ/2=1+cosθ/2, tan²θ/2=1-cosθ/1+cosθ sinαcosβ=1/2{sin(α+β)+sin(α-β)} cosαcosβ=1/2{cos(α+β)+cos(α-β)} sinαsinβ=-1/2{cos(α+β)-cos(α-β)} sinα+sinβ=2sin(α+β/2)cos(α-β/2) sinα-sinβ=2cos(α+β/2)sin(α-β/2) cosα+cosβ=2cos(α+β/2)cos(α-β/2) cosα-cosβ=-2sin(α+β/2)sin(α-β/2) }}}