The original problem is f(x)=sinxcosx
to find the derivative:
f(x) = sinx(-six) + cosx(cosx)
= -sin²x +cos²x
but could someone explain to me how -sin²x +cos²x becomes
cos2x?
2007-02-21 00:51:51 · 5 answers · asked by sciencenerd 2 in Science & Mathematics ➔ Mathematics
There are two ways
1) the easiest one sin(2x) = 2sinx cosx so 1/2 sin2x = sin*cos and your derivative is cos2x
Derivating as a product
=sinx *(-sinx) +cos^2x = cos^2x= sin^2x= cos 2x because
cos(2x) =cos(x+x) cosx*cosx -sinx * sin x =cos^2x-sin^2x
2007-02-21 07:11:35 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
also cos^2 and sin^2 = 1
btw, how do you get the exponents in your text?
2007-02-21 01:10:42 · answer #2 · answered by minorchord2000 6 · 0⤊ 1⤋
cos 2x= cos (x + x)
= cos x cos x - sin x sin x
=cos²x - sin²x
=-sin²x + cos²x
2007-02-21 01:17:13 · answer #3 · answered by usp 2 · 1⤊ 0⤋
That's the double angle formula.
2007-02-21 00:58:22 · answer #4 · answered by Anonymous · 1⤊ 0⤋
aliter:
using sin2x = 2 sinx cosx
f(x)=sinxcosx = (sin2x)/2
f '(x) = cos2x
2007-02-21 04:48:40 · answer #5 · answered by qwert 5 · 0⤊ 0⤋