document.write( "Question 32084: Venugopalramana has kindly calculated and ploted the data from y=3cos(2t+ 40 degrees),0 < t < 360 degrees. But I do not know how the data was calculated and how he plotted the data. Could Venugopalramana or another tutor please explain the steps to calculating the data. Thankkyou \n" ); document.write( "
Algebra.Com's Answer #18625 by venugopalramana(3286)\"\" \"About 
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sketch y=3cos(2t+ 40 degrees),0 < t < 360 degrees
\n" ); document.write( "CALCULATE THE DATA AS SHOWN BELOW AND PLOT THEM ON A GRAPH.IT WILL
\n" ); document.write( "LOOK AS GIVEN BELOW.
\n" ); document.write( "T-DEG. T-RADIANS y=3cos(2t+ 40 degrees)
\n" ); document.write( "0 0.00 2.30........OK I WILL SHOW ONE EXAMPLE.LET US DO FOR T=30 DEGREES
\n" ); document.write( "WE CAN CALCULATE WITH DEGREES IF YOU HAVE CALCULATOR WITH DEGREE
\n" ); document.write( "MEASURE.....FIND 2T+40=2*30+40=60+40=100...NOW FIND COS(100) DEGREES IN
\n" ); document.write( "CALCULATOR..YOU WILL GET IT AS -0.1733....MULTIPLY WITH 3 TO GET
\n" ); document.write( "-0.52...CONTINUE THIS WAY FOR DIFFERENT VALUES OF T STARTING FROM 0
\n" ); document.write( "DEGREES AND GOING IN STEPS OF 30 OR 60 DEGREES AS YOU LIKE.
\n" ); document.write( "AFTER TABULATING AS SHOWN HERE TAKE A SUITABLE SCALE FOR X AND Y AXIS
\n" ); document.write( "AND PLOT THEM FOR T=0 TO 360 DEGREES AS DESIRED....VENUGOPAL
\n" ); document.write( "30 0.52 -0.52
\n" ); document.write( "60 1.05 -2.82
\n" ); document.write( "90 1.57 -2.30
\n" ); document.write( "120 2.09 0.52
\n" ); document.write( "150 2.62 2.82
\n" ); document.write( "180 3.14 2.30
\n" ); document.write( "210 3.67 -0.52
\n" ); document.write( "240 4.19 -2.82
\n" ); document.write( "270 4.71 -2.30
\n" ); document.write( "300 5.24 0.52
\n" ); document.write( "330 5.76 2.82
\n" ); document.write( "360 6.28 2.30
\n" ); document.write( "
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