How do you find the point (x,y) on the unit circle that corresponds to the real number #t=(11pi)/6#? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer jess_r Feb 7, 2017 (#(sqrt3)/2#, -#1/2#) Explanation: #(11pi)/6# corresponds to 330 degrees on the unit circle, which also has the same point values as #pi/6#, but the y value is negative because it is in the 4th quadrant. 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 5379 views around the world You can reuse this answer Creative Commons License