The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
This is a perfect time to use l'Hospital's rule: If a fraction becomes 0/0 at the limit (like this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator)/(derivative of the denominator) . I'm not sure why l'Hospital's rule stuck with me all these years. But when it's appropriate, like for this one, you can't beat it.
a formula is a formula <3 simples!
According to de Moivre's formula, cos3x + isin3x = (cosx + isinx)3 = cos3x + 3cos2x*isinx + 3cosx*i2sin2x + i3sin3x Comparing the imaginary parts, isin3x = 3cos2x*isinx + i3sin3x so that sin3x = 3cos2x*sinx - sin3x = 3*(1-sin2x)sinx - sin3x = 3sinx - 4sin3x
int cos3x=sin3x/3+c
S sin3X dx =-1/3 cos3X
The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
If you want sin(3x) + cos(3x) = 6, then this is impossible. Sine and cosine will only return values between -1 and 1, so the expression sin(3x) + cos(3x) could only take values from -2 to 2, although even this is to great as sine and cosine of the same number will never both be 1 or -1. Similarly, if you want a solution to sin3x + cos3x = 6, then this is also impossible, because any power of a number between -1 and 1 will itself be between -1 and 1.
A proof in calculus is when it will make a statement, such as: If y=cos3x, then y'''=18sin3x. Then it will tell you to do a proof. This means you have to solve the equation step by step, coming to the solution, which should be the same as in the statement. If you do come to the same answer as in the statement, then you just correctly did a calculus proof.
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
What Is The Formula What Is The Formula What Is The Formula For Calcium Phosphate ?
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