Trigonometry

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Submitted By veanz13
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sin(theta) = a / c | csc(theta) = 1 / sin(theta) = c / a | cos(theta) = b / c | sec(theta) = 1 / cos(theta) = c / b | tan(theta) = sin(theta) / cos(theta) = a / b | cot(theta) = 1/ tan(theta) = b / a |

sin(-x) = -sin(x) csc(-x) = -csc(x) cos(-x) = cos(x) sec(-x) = sec(x) tan(-x) = -tan(x) cot(-x) = -cot(x) sin2(x) + cos2(x) = 1 | tan2(x) + 1 = sec2(x) | cot2(x) + 1 = csc2(x) | sin(x y) = sin x cos y cos x sin y | | cos(x y) = cos x cosy sin x sin y | | tan(x y) = (tan x tan y) / (1 tan x tan y) sin(2x) = 2 sin x cos x cos(2x) = cos2(x) - sin2(x) = 2 cos2(x) - 1 = 1 - 2 sin2(x) tan(2x) = 2 tan(x) / (1 - tan2(x)) sin2(x) = 1/2 - 1/2 cos(2x) cos2(x) = 1/2 + 1/2 cos(2x) sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 ) cos x - cos y = -2 sin( (x-y)/2 ) sin( (x + y)/2 ) Trig Table of Common Angles | angle | 0 | 30 | 45 | 60 | 90 | sin2(a) | 0/4 | 1/4 | 2/4 | 3/4 | 4/4 | cos2(a) | 4/4 | 3/4 | 2/4 | 1/4 | 0/4 | tan2(a) | 0/4 | 1/3 | 2/2 | 3/1 | 4/0 |

Given Triangle abc, with angles A,B,C; a is opposite to A, b oppositite B, c opposite C: a/sin(A) = b/sin(B) = c/sin(C) (Law of Sines) c2 = a2 + b2 - 2ab cos(C)b2 = a2 + c2 - 2ac cos(B)a2 = b2 + c2 - 2bc cos(A) | | (Law of Cosines) |
(a - b)/(a + b) = tan 1/2(A-B) / tan 1/2(A+B) (Law of…...

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