If #cosa=m#, what is #sin(a-pi/2)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Shwetank Mauria Feb 2, 2017 #sin(a-pi/2)=-m# Explanation: We can use two formulas here. One #sin(-A)=-sinA# and #sin(pi/2-A)=cosA# Hence #sin(a-pi/2)=sin(-(pi/2-a))# = #-sin(pi/2-a)=-cosa=-m# 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 3110 views around the world You can reuse this answer Creative Commons License