J jbreu New member Joined Jan 28, 2011 Messages 6 Jan 28, 2011 #1 3x + 3xcos(3x) / 8sin(3x)cos(3x) = 3x / 8sin(3x)cos(3x) + 3x / 8sin(3x) I understand how the numerator simplifies, but why is cos(3x) simplified from the denominator in the second fraction?
3x + 3xcos(3x) / 8sin(3x)cos(3x) = 3x / 8sin(3x)cos(3x) + 3x / 8sin(3x) I understand how the numerator simplifies, but why is cos(3x) simplified from the denominator in the second fraction?
D Deleted member 4993 Guest Jan 28, 2011 #2 jbreu said: 3x + 3xcos(3x) / 8sin(3x)cos(3x) = 3x / 8sin(3x)cos(3x) + 3x / 8sin(3x) I understand how the numerator simplifies, but why is cos(3x) simplified from the denominator in the second fraction? Click to expand... \(\displaystyle \frac{3*x + 3*x*cos(3*x)}{8*sin(3*x)*cos(3*x)} \ =\) \(\displaystyle \ \frac{3*x}{8*sin(3*x)*cos(3*x)} + \frac{3*x*cos(3*x)}{8*sin(3*x)*cos(3*x)} \ =\) \(\displaystyle \ \frac{3*x}{8*sin(3*x)*cos(3*x)} + \frac{3*x}{8*sin(3*x)} \ =\)
jbreu said: 3x + 3xcos(3x) / 8sin(3x)cos(3x) = 3x / 8sin(3x)cos(3x) + 3x / 8sin(3x) I understand how the numerator simplifies, but why is cos(3x) simplified from the denominator in the second fraction? Click to expand... \(\displaystyle \frac{3*x + 3*x*cos(3*x)}{8*sin(3*x)*cos(3*x)} \ =\) \(\displaystyle \ \frac{3*x}{8*sin(3*x)*cos(3*x)} + \frac{3*x*cos(3*x)}{8*sin(3*x)*cos(3*x)} \ =\) \(\displaystyle \ \frac{3*x}{8*sin(3*x)*cos(3*x)} + \frac{3*x}{8*sin(3*x)} \ =\)
