How do you evaluate the six trigonometric functions given t=−0.56? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer sankarankalyanam Feb 20, 2018 As calculated below Explanation: Given #t = -0.56^c# #sin t = sin (-0.56) = - 0.5312# #cos t = cos (-0.56) = 0.8473# #tan t = tan (-0.56) = - 0.6269# #csc t = 1 / sin (-0.56) = -1/0.5312 = -1.8825# #sec t = 1/cos (-.56) = 1/0.8473 = 1.1802# #cot t = 1 / tan t = 1 / -0.6269 = -1.5952# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 1507 views around the world You can reuse this answer Creative Commons License