Anti-Differentiation involving sin and cos?

You are correct:

COS will always be NEGATIVE
and SIN will always be POSITIVE

This is what you got and you are correct. I think it will be easier to come up with 127th if we look at the fraction

Let N = nth derivative

notice if N = even, we have positive sin 2x and if N = odd, we have negative cos 2x

We have this pattern:

if N = even: (1/2)^(N-2)*sin (2x)
if N = odd: - (1/2)^(N - 2)*cos (2x)

N = 127: - (1/2)^(127-2)*cos (2x)

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Lets see your anti-derivative

N = 2, even: sin2x = (1/2)^(2-2)sin 2x

N = 3, odd : -0.5cos2x = - (1/2) ^(3 - 2)cos 2x

4) 0.25sin2x = (1/2)^(4 - 2)sin 2x = (1/2)*(1/2) sin2x

5) -0.125cos2x = - (1/2)^(5 - 2) cos 2x

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I forgot to say that N starts at 2 for this pattern

1a. v = 10 = 6+ 8sin ½t 4 = 8sin ½t ½ = sin ½t arcsin ½ = ½t t = 2arcsin ½ t = 2(?/6) = ?/3 b. a = dv/dt = 4 cos(½t) a(0) = 4 cos 0 = 4(a million) = 4 2. you have to be greater sparkling with your fact of the challenge via using parentheses. There are 3 obtainable approaches to interpret what you have written above.

You are incorrect, cos will not always be negative and sin will not always be positive.

You should get something that follows this cycle:

sin antidifferentiates to -cos
-cos antidifferentiates to -sin
-sin antidifferentiates to cos
cos antidifferentiates to sin
sin antidifferentiates to -cos

and so on.

f(x) = 4sin2x
1).-2cos2x
2) -sin2x
3). 0.5cos2x
4) 0.25sin2x
5) -0.125cos2x