Evaluate tan(arccos(-1/4))

Question

My teacher taught us how to solve equations like y = cos(x) but not expressions

Answer 1

Answer 2

tan(arccos(-1/4))

= sqr(4^2-1)/-1

= -sqr15

= about -3.873

the angle is in quadrant II = about 104.478 degrees=about 1.823 radians

construct a right triangle with hypotenuse = 4, adjacent side =-1

opposite side = sqr(4^2 - (-1)^2) = sqr15

tangent of the angle = sqr15/-1 =-sqr15 = opposite side/adjacent side

usually the math convention is to restrict the domain of arccosx to the 1st and 2nd quadrants, with 0<x<pi

that eliminates a 2nd possible solution tan(arccos(-1/4)=+sqr15

leaving only -sqr15 as the solution