inverse trig functions
posted by abby .
evaluate the following expressions:
i know the answers.. i jus don't know how to solve them =( PLEASE help me
I assume your sec^-1 notation denot4es the arcsecant function, etc.
sec^-1 (5/3) = cos^-1 (3/5). Think of a 3,4,5 right triangle. The tangent of the angle is 4/3.
sec^-1 (25/7) = cos^-1 (7/25)
Think of a 7,25,24 right triangle. The hypotenuse is 25 and the adjacent side is 7. The tangent is 24/7
csc^(5/3) = sin^-1 (3/5). Think of a 3,4,5 right triangle again. The hypotenuse is 5 and the opposite side of the angle in question is 3. The cotangent is (adjacent side)/(opposite side) = 4/3.