How do you find cos 2x, given #sinx=1/5# and cos x <0? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Nghi N Apr 25, 2017 #cos 2x = 23/25# Explanation: Use trig identity: #cos 2a = 1 - 2sin^2 a# In this case: #cos 2x = 1 - 2/25 = 23/25# 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 7421 views around the world You can reuse this answer Creative Commons License