If sinx = -3/5, x is in quadrant III, then how do you find sin2x? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer Gerardina C. Jun 5, 2016 #sin2x=24/25# Explanation: #sin2x=2sinxcosx# #cosx=-sqrt(1-sin^2x# (cosx <0 in quadrant III) so #cosx=-sqrt(1-9/25)# that's #cosx=-4/5# and #sin2x=2(-3/5)(-4/5)# that's #24/25# 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 6424 views around the world You can reuse this answer Creative Commons License