If #sin x= 4/5#, how do you find cos x? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer bp Apr 16, 2015 cosx= #3/5# Use Trignometrical identity cosx = #sqrt(1-sin^2 x)# cos x = #sqrt(1 -16/25)# =#sqrt(9/25)# = #3/5# to be the value in the first quadranr. 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 65748 views around the world You can reuse this answer Creative Commons License