If #sinx=1/4# and #tanx# is positive, find #cosx#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Shwetank Mauria Mar 1, 2017 #cosx=sqrt15/4# Explanation: As #sinx# and #tanx# both are positive, #x# lies in Quadrant 1 and #cosx=sinx/tanx# too is positive. Hence, #cosx=sqrt(1-sin^2x)=sqrt(1-(1/4)^2)=-sqrt(1-1/16)=-sqrt(15/16)=sqrt15/4# 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 4995 views around the world You can reuse this answer Creative Commons License