kasse Messages 383 Reaction score 1 Thread starter Jan 14, 2007 #1 In my book, (cos4x)^2 is written 1+cos8x without referring to any formula. Which trig. identity is used here?
In my book, (cos4x)^2 is written 1+cos8x without referring to any formula. Which trig. identity is used here?
cristo Staff Emeritus Science Advisor Messages 8,145 Reaction score 75 Jan 14, 2007 #2 Try looking at the identity for cos(2x)
arunbg Messages 594 Reaction score 0 Jan 14, 2007 #3 The correct identity is (cos4x)^2 = (1+cos8x)/2 .
kasse Messages 383 Reaction score 1 Jan 14, 2007 #4 cristo said: Try looking at the identity for cos(2x) You mean cos(2x) = (cosx)^2 - (sinx)^2 ?
cristo Staff Emeritus Science Advisor Messages 8,145 Reaction score 75 Jan 14, 2007 #5 kasse said: You mean cos(2x) = (cosx)^2 - (sinx)^2 ? Yes, and as arunbg says, there is a factor of 1/2 missing from your given identity.
kasse said: You mean cos(2x) = (cosx)^2 - (sinx)^2 ? Yes, and as arunbg says, there is a factor of 1/2 missing from your given identity.
cristo Staff Emeritus Science Advisor Messages 8,145 Reaction score 75 Jan 14, 2007 #8 JJ420 said: the identity is cos^2x = (1 + cos2x)/2 is it not? One can derive this from the double angle identity for cos(2x) using further the identity that cos2x+sin2x=1
JJ420 said: the identity is cos^2x = (1 + cos2x)/2 is it not? One can derive this from the double angle identity for cos(2x) using further the identity that cos2x+sin2x=1
dextercioby Science Advisor Insights Author Messages 13,423 Reaction score 4,244 Jan 15, 2007 #10 Nope. [tex]\sin^{2} x=\frac{1-\cos 2x}{2}[/tex] Daniel.