Title: Derive trigonometric identity? Post by: Tom291 on Jul 25, 2012 Derive the trigonometric identity: sin^2x + cos^x=1
using this equality: cox(x+y) = cosxcosy - sinxsiny I'm not real sure where to even begin, so any help would be great. Thank you! Title: Derive trigonometric identity? Post by: 0248 on Jul 25, 2012 Consider cos(x+(-x)) = cos (x-x) = cos 0 = 1.
Using your formula, cos (x+(-x)) = cos(x)cos(-x) - sin(x)sin(-x). Since cos(x) is an even function, cos(-x) = cos(x). Since sin(x) is an odd function, sin(-x) = - sin(x). Thus you have cos (x-x) = cos(x)cos(x) + sin(x)sin(x) cos (0) = 1 = cos^2(x) + sin^2(x). Title: Derive trigonometric identity? Post by: Abdelrhaman on Jul 26, 2012 cos(x+y)= cosx cosy-sinx siny.
Put y=-x; cos 0=cosx cosx - sinx [-sinx] 1= cos^2x+sin^2x. QED. Title: Derive trigonometric identity? Post by: fishnchips on Jul 26, 2012 First, before I begin working on the approach suggested,
Pick any point on a circle inscribed about the origin of the X-Y plane. By Pythagoras we know that the square of the radius of the circle is equal to the sum of the square of the x coordinate of the point and the square of the y coordinate. or (x/r)² + (y/r)² = 1r² = x² + y² if you multiply both sides by 1/r² you get r² /r² = x²'/r/² + y²/r² or (x/r)² + (y/r)² = 1 Measuring the angle (?) from the positive x axis, The trigonometric functions sine and cosine are defined as y/r and x/r respectively, so sin(?)=y/r, and cos(?)=x/r Plugging these back into (x/r)² + (y/r)² = 1, we get... cos²(?) + sin²(?) = 1 That's the proof with which I'm familiar. And the derivation of the sin(a+b) and cos(a+b) functions are geometric and impossible to do on this site. That said, I'll spend a little time working on the problem you have presented. Ok, this is as far as I can get it. cox(x+y) = cosxcosy - sinxsiny If x=y then you have cox(x+y)=cos(2x) = cosxcos x- sinxsinx=cos²(x) - sin²(x) But, cos(2x) is not necessarily = 1. Give the points to Derek... He's right.;.. and first... I hadn't thought of that. |