# Math differentiation rearrangementWatch

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#1
(2sintcost) / 2-2(1-sin^2t) = Anyone know how I go about doing this
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5 years ago
#2
Is this with respect to t?
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5 years ago
#3
(Original post by REASON_Lighters)
(2sintcost) / 2-2(1-sin^2t) = Anyone know how I go about doing this
isnt 2-2(1-sin^2t) just 2sin^2t? and you can take it from there
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#4
(Original post by physics4ever)
isnt 2-2(1-sin^2t) just 2sin^2t? and you can take it from there
thank you, i get cot t in the end which is right. How do you know that thought - like are you supposed to know that or is there some algebraic manipulation you have to do.

Thanks.
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#5
bump
0
5 years ago
#6
(Original post by REASON_Lighters)
thank you, i get cot t in the end which is right. How do you know that thought - like are you supposed to know that or is there some algebraic manipulation you have to do.

Thanks.
2-2(1-sin^2t) = (2-2)-2(-sin^2t) = -2(-sin^2t) = 2sin^2t

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#7
(Original post by Phichi)
2-2(1-sin^2t) = (2-2)-2(-sin^2t) = -2(-sin^2t) = 2sin^2t

SORRY, But I dont get the 2-2-2 part, thanks
0
5 years ago
#8
(Original post by REASON_Lighters)
thank you, i get cot t in the end which is right. How do you know that thought - like are you supposed to know that or is there some algebraic manipulation you have to do.

Thanks.
expand 2(1-sin^2t) so 2 - 2(1-sin^2t) = 2 - (2-2sin^2t) = 2 - 2 + 2sin^2t = 2sin^2t
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#9
(Original post by physics4ever)
expand 2(1-sin^2t) so 2 - 2(1-sin^2t) = 2 - (2-2sin^2t) = 2 - 2 + 2sin^2t = 2sin^2t
I love you. Thank you so much, you are a hero x)
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5 years ago
#10
(Original post by REASON_Lighters)
I love you. Thank you so much, you are a hero x)
Spoiler:
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no problem
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