Question

Find arccos[cos(7pi/2)] please show work. I will mark brainliest

Answer 1

Arccos(cos 7pi/2) 
cos 7pi/2 = cos 630 = cos (630-360) = cos 270 = 0 
arccos(0) = 90 since cos 90 = cos 270 
(x): 0 - 30 - 45 - 60 - 90 
sin (x) 0 - 1/2 - 1/2 √3 - 1/2 √3 - 1 
cos (x) 1 - 1/2 √3 - 1/2 √3 - 1/2 - 0 
tan (x) 0 - 1/√3 - 1 - √3 - ∞ 

Rules for sin cos tan value in each quadrant : 
quadrant = I(0-90) - II (90-180)- III(180-270) - IV(270-360) 
sin (x) = positive- positive - negative - negative 
cos (x) = positive- negative- negative - positive 
tan (x) = positive- negative- positive - negative

Answer 2

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