How do you find cos 2x, given cot x = 3/5 and csc x < 0? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Vinícius Ferraz Dec 5, 2015 # - 8/17# Explanation: #cos x / sin x = 3/5# #(1 - sin^2 x) / sin^2 x = 9/25# #9 sin^2 x = 25 - 25 sin^2 x# #sin x = ± sqrt{ 25/34}# #1/sin x < 0 Rightarrow sin x = - 5/ sqrt 34# #cos 2x = (1 - sin^2 x) - sin^2 x = 1 - 2 * 25 / 34# Answer link Related questions What does it mean to find the sign of a trigonometric function and how do you find it? What are the reciprocal identities of trigonometric functions? What are the quotient identities for a trigonometric functions? What are the cofunction identities and reflection properties for trigonometric functions? What is the pythagorean identity? If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot 325^@# lie in and what is the sign? How do you use use quotient identities to explain why the tangent and cotangent function have... How do you show that #1+tan^2 theta = sec ^2 theta#? See all questions in Relating Trigonometric Functions Impact of this question 4903 views around the world You can reuse this answer Creative Commons License