Asked by Julie
Differentiate
A) y = -Cos2x
B) y = Sin2tetta - 2Cos2tetta
C) f(tetta) = negative pie*Sin(2tetta - pie)
Tetta means that zero with a line going horizontally across mid zero. Like In trigonometric identities you have ___ tetta....
A) y = -Cos2x
B) y = Sin2tetta - 2Cos2tetta
C) f(tetta) = negative pie*Sin(2tetta - pie)
Tetta means that zero with a line going horizontally across mid zero. Like In trigonometric identities you have ___ tetta....
Answers
Answered by
Bosnian
Go on: wolframalpha dot com
When page be open in rectangle type:
derivative cos(2x)
and click option =
After few seconds when you see result click option:
Show steps
Then in in rectangle type:
derivative sin(2theta)-2cos(2theta)
and click option =
then Show steps
C)
sin ( theta - pi ) = - sin ( theta )
sin ( 2 theta - pi ) = - sin ( 2 theta )
f(theta) = - pi * sin( 2theta - pi ) =
- pi * [ -sin ( 2 theta ) ]=
pi * sin ( 2 theta )
When page be open in rectangle type:
derivative cos(2x)
and click option =
After few seconds when you see result click option:
Show steps
Then in in rectangle type:
derivative sin(2theta)-2cos(2theta)
and click option =
then Show steps
C)
sin ( theta - pi ) = - sin ( theta )
sin ( 2 theta - pi ) = - sin ( 2 theta )
f(theta) = - pi * sin( 2theta - pi ) =
- pi * [ -sin ( 2 theta ) ]=
pi * sin ( 2 theta )
Answered by
Julie
A) y'= 2sin(2x)
B) y'= 4sin(2theta) -2cos(2theta)
C) y'= 2piCos(2theta)
B) y'= 4sin(2theta) -2cos(2theta)
C) y'= 2piCos(2theta)
Answered by
Bosnian
A) y'= - 2sin(2x)
B) y'= 4 sin ( 2 theta ) + 2 cos ( 2theta ) =
2 [ 2 sin (2 theta ) + cos ( theta ) ]
C) y'= 2piCos(2theta)
B) y'= 4 sin ( 2 theta ) + 2 cos ( 2theta ) =
2 [ 2 sin (2 theta ) + cos ( theta ) ]
C) y'= 2piCos(2theta)
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