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[quote][i]Originally posted by Anand[/i] [br]x= acos[theta]+bsin[theta] y= asin[theta]bcos[theta] prove that X[squared]+y[squared] = a[squared]+b[squared] so x[squared]+y[squared]= a[squared]cos[squared][theta]+b[squared]sin[squared][theta]+2acos[theta]bsin[theta]+a[squared]sin[squared][theta]+b[squared]cos[squared][theta]2asin[theta]bcos[theta] =a[squared](sin[squared][theta]+cos[squared][theta])+b[squared](sin[squared][theta]+cos[squared][theta])+2[acos[theta]bsin[theta]asin[theta]bcos[theta]] =a[squared](1)+b[squared](1)+2 [acos[theta]bsin[theta]asin[theta]bcos[theta] ] What to do after this step ? [/quote]
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T O P I C R E V I E W
Anand
Posted  08/02/2011 : 02:57:19
x= acos
+bsin
y= asin
bcos
prove that X
+y
= a
+b
so x
+y
=
a
cos
+b
sin
+2acos
bsin
+a
sin
+b
cos
2asin
bcos
=a
(sin
+cos
)+b
(sin
+cos
)+2[acos
bsin
asin
bcos
]
=a
(1)+b
(1)+2 [acos
bsin
asin
bcos
]
What to do after this step ?
2 L A T E S T R E P L I E S (Newest First)
Anand
Posted  08/08/2011 : 04:09:59
Got it
Thanks
royhaas
Posted  08/02/2011 : 07:23:42
The term in brackets is zero.
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