How do you evaluate #4 cos^2(-pi/2)#? Trigonometry Right Triangles Basic Trigonometric Functions 1 Answer Oscar L. Aug 4, 2016 As #cos(-\pi/2)=0#, #4cos^2(-pi/2)=4(0^2)=0#. Explanation: Remember that any multiple of #\pi# plus or minus #\pi/2# gives a cosine equal to zero. Just a multiple of #\pi# gives a sine equal to zero. Answer link Related questions What are the 6 basic trigonometric functions? How are the 6 basic trigonometric functions related to right triangles? How are the 6 basic trigonometric functions related to the Unit Circle? What is Sine, Cosine and Tangent? How do you find the Sine, Cosine, or Tangent of an angle? Can you use trigonometric functions for any kind of triangle? For a right triangle ABC, how do you find the sine, cosine and tangent of angle A? If #sin A = 3/5#, what is #sec A#? If #sin A= 1/(sqrt(2))#, what is #cot A/csc A#? How do you write #(sec theta - 1)(sec theta + 1)# in terms of sine and cosine? See all questions in Basic Trigonometric Functions Impact of this question 13435 views around the world You can reuse this answer Creative Commons License