the primitive function is the anti-derivative. so cosx becomes sinx and sinx becomes -cosx?
s(t)=-cos3t+sin3t + C
Now I got it. I´d gotten confused between the derivative and the anti-derivative. Thanks for confirming that I´d made the mistake!
Homework Statement
Find the primitive function s(t) for v(t)=3(sin(3t) + cos(3t)) for which s(0)=0
The Attempt at a Solution
v(t) =3(sin(3t) + cos(3t)) = 3sin3t +3cos3t
s(t) =cos3t -sin3t + C
0 = cos3(0) - sin3(0) + C
0= 1 + C
C=-1
such that my primitive function is cos3t -...
Homework Statement
According to my math book, I solved the following trig equation correctly:
1)
cos3x=0,500=
3x = 60° + n360°
x=20+n120°
2)I also solved this problem correctly:
4sin^2x -3sin^x= 0
sinx(4sinx-3)=0
x=n*360°
or
4sinx=3
sinx=3/4
x= ca. 49°
x=49° + n360° or 131+ n360°.
Now I´m...
Somehow it keeps on getting more and more complicated. Do you have any idea on how I could keep it "simple"?
sin4x + 2sin2x = 8sinxcos^3x
on the left hand side:
2sin2xcos2x + 2sin2x =
2*2sinxcosx(cos^2x - sin^2x) + 2*2sinxcosx =
4sinxcos^3x - 2sin^3xcosx + 4sinxcosx
It really doesn´t feel right...
Ok, did you derive that from sin2x = 2sinxcosx?
In that case, I don´t understand why sin4x = 2sin2xcos2x and not 4sin2xcos2x (i.e twice as much as sin2x = 2sinxcosx?)
Homework Statement
show that sin4x + 2sin2x = 8sinxcos^3x
Homework Equations
sin2x =2sinxcosx
The Attempt at a Solution
I started out by letting sin4x = sin(2x*2) so that I could plugg in sin2x = 2sinxcosx in the equation.
sin(2x*2) +2sin2x =
sin2*sin2x +2sin2x =
sin2 *...