Prove cos x(1-cos2x) = sin x(sin2x)?

Answer 1

= cos[x] ( 1- (1-2sin²[x]) )
= cos[x] (2sin²[x])
= sin[x] (2sin[x] cos[x])
= sin[x] sin[2x]


©

Answer 2

LHS
cos x [1 - (1 - 2 sin ² x) ]
cos x [ 2 sin ² x ]
sin x (2 sin x cos x)
sin x sin 2x

RHS
sin x sin 2x

LHS = RHS

Answer 3

ò(1-sin x)/(1+sin x) dx
= ò(1-sin x)2/(1-sin2x) dx
= ò(1 - 2 sin x + sin2x)/cos2x dx
= òsec2x dx - ò2 sin x/cos2x dx + òtan2x dx
= òsec2x dx - ò2 sin x/cos2x dx + ò(sec2x -1) dx
= tan x - 2 sec x + (tan(x)-x) + C
= 2 tan(x) - 2 sec(x) - x + C

This brings up another interesting identity

sec(x) - tan(x) = (1-tan(x/2))/(1+tan(x/2))

Proof: (1+tan²(x/2))/(1-tan²(x/2)) - 2 tan(x/2)/(1-tan²(x/2)), from the Weierstrass t-substitutions,
= (1-tan(x/2))²/(1-tan²(x/2))
= (1-tan(x/2)/(1+tan(x/2))

Answer 4

use the sum formula for sine sin(90 +@) = sin90cos@ +sin@cos90 = (1)cos@ +sin@ (0) = cos@

Answer 5

1-cos(2x)=2 sin^2(x)
cos x(1-cos(2x) )=cos(x)*2*sin^(2)x
=cos(x)*2*sin(x)*sin(x)
2sin(x)*cos(x)=sin(2x)
using this in the above equation,
=sin(x)*sin(2x)

Answer 6

take L.H.S cosx(2 sinx*sinx) =2(sinx)(sinx)(cosx)
(bcoz of formula 1-cos2x=2 sinx*sinx)
now take R.H.S sinx(2 sinx*cosx) (FORMULA=SIN2x=2 sinx cosx)
=2(sinx)(sinx)(cosx)
therefore L.H.S=R.H.S