How do you find the important points to graph #y = (1/2)sin(x - pi)#?

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Answer 1 · 2016-05-01T01:31:28

You can use the given trigonometric equation to compute for the rectangular coordinate points #(x, y)#, then use these points to draw your graph.

From the given equation
#y=1/2*sin(x-pi)#

Compute the rectangular coordinate points #(x, y)#

#x# in degrees and radians

#"degrees x y " " " " y in decimals"#
#" "0##" "0##" "0##" "0#

#" "30##" "pi/6##" "-1/4##" "-0.25#

#" "45##" "pi/4##" "(-sqrt2)/4##" "-0.3535533906#

#" "60##" "pi/3##" "(-sqrt3)/4##" "-0.4330127019#

#" "90##" "pi/2##" "-1/2##" "-0.5#

#" "120##" "(2pi)/3##" "(-sqrt3)/4##" "-0.4330127019#

#" "135##" "(3pi)/4##" "(-sqrt2)/4##" "-0.3535533906#

#" "150##" "(5pi)/6##" "-1/4##" "-0.25#

#" "180##" "pi##" "0##" "0#

#" "210##" "(7pi)/6##" "1/4##" "0.25#

#" "225##" "(5pi)/4##" "(sqrt2)/4##" "0.3535533906#

#" "240##" "(4pi)/3##" "(sqrt3)/4##" "0.4330127019#

#" "270##" "(3pi)/2##" "1/2##" "0.5#

#" "300##" "(5pi)/3##" "(sqrt3)/4##" "0.4330127019#

#" "315##" "(7pi)/4##" "(sqrt2)/4##" "0.3535533906#

#" "330##" "(11pi)/6##" "1/4##" "0.25#

#" "360##" "2pi##" "0##" "0#

The draw your graph

graph{(y-1/2*sin(x-pi))=0[-5,5,-2.5,2.5]}

God bless....I hope the explanation is useful.