Trigonometry Cheatsheet Version 2.1 © 2008 Ankur Banerjee * sin (A+B) * cos (A+B) = = sinA cosB + cosA sinB cosA cosB – sinA sinB 2 2 2 2 = = = = cos (A+B) + cos (A–B) cos (A–B) – cos (A+B) sin (A+B) + sin (A–B) sin (A+B) – sin (A–B) cosA cosB sinA sinB sinA cosB cosA sinB sin(A+B) sin(A–B) cos(A+B) cos(A–B) = = sin2A – sin2B cos2A – sin2B sinA + sinB sinA – sinB cosA + cosB cosA – cosB = = = = 2 sin[(A+B) / 2] cos[(A–B) / 2] 2 cos[(A+B) / 2] sin[(A–B) / 2] 2 cos[(A+B) / 2] cos[(A–B) / 2] –2 sin[(A+B) / 2] sin[(A–B) / 2] * tan (A+B) = tanA + tanB 1 – tanA tanB * cot (A+B) = cotA cotB – 1 cotB + cotA sin 2A = 2 sinA cosA cos 2A = cos2A – sin2A = tan 2A = 2 tanA 1 – tan2A 2 sin2A 2 cos2A = = 1 – cos2A 1 + cos2A = 2 tanA 1 + tan2A 1 – tan2A 1 + tan2A * Substitute '+' with '−' to get corresponding formula www.ankurb.info tan2(A/2) = 1 – cosA 1 + cosA sin 3A = = 3 sinA – 4 sin3A 4 sinA sin(60° – A) sin(60° + A) cos 3A = = 4 cos3A – 3 cosA 4 cosA cos(60° – A) cos(60° + A) tan 3A = 3 tanA – tan3A 1 – 3 tan2A tanA tan(60° – A) tan(60° + A) = General Solutions For sinA = k; A = nπ + (–1)n B For cosA = k; A = 2nπ ± B For tanA= k; A = nπ + B For sin2A = sin2B, cos2A = cos2B, tan2A = tan2B; A = nπ ± B sinA + sin(A+B) + sin(A+2B) + ... + sin(A + (n–1)B) = [ sin(A + (n–1)B/2) sin(nB/2) ] / sin(B/2) cosA + cos(A+B) + cos(A+2B) + ... + cos(A + (n–1)B) = [ cos(A + (n–1)B/2) sin(nB/2) ] / sin(B/2) In a triangle ABC ∑ sin A = 4 ∏ cos(A/2) ∑ cos A = 1 + 4 ∏ sin(A/2) ∑ sin 2A = 4 ∏ sinA ∑ cos 2A = –1 – 4∏ cosA ∑ tan A = ∏ tanA ∑ tan(A/2) tan(B/2) = 1 ∑ cotA cotB = 1 ∑ cosA + cos(∑A) = 4 ∏ cos[(A+B)/2] ∑ sinA – sin(∑A) = 4 ∏ sin[(A+B)/2 ] tan (∑A) = ∑ tanA – ∏ tanA 1 – ∑ tanA tanB Sine Rule: Cosine Rule: Projection Rule: (sinA / a) = (sinB / b) = (sinC / c) a2 = b2 + c2 + 2 bc cosA a = b cosC + c cosB www.ankurb.info