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

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Question:

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

Answers

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

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