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#1
Prove that for all values of x

(2 tan x) - (sin 2x) = 2sin(squared)xtan x

thanks!
0
#2
anyone?
0
17 years ago
#3
RHS = 2sin^2(x)tan(x) = 2(1-cos^2(x))tan(x) = 2(tan(x) - cos(x)sin(x)) = 2tan(x) - 2sin(x)cos(x) = 2tan(x) - sin(2x) = LHS
0
17 years ago
#4
(Original post by Kurdt Morelo)
Prove that for all values of x

(2 tan x) - (sin 2x) = 2sin(squared)xtan x

thanks!
2 tan x - Sin 2x = 2 Sin ^2 x . tan x
2 Sin x/Cos x - 2 Sin x Cos x = 2 Sin ^2 x . (Sin x/ Cos x)
2 Sin x .[(1/cos x) - cos x ] = 2 Sin x .(Sin ^2 x / Cos x)
[(1-cos ^ 2x)/cos x] = Sin ^2 x / Cos x
cos ^2 x = 1 - Sin ^2 x
So :
[(1-1+Sin ^2 x)/cos x] = Sin ^2 x/ Cos x
Sin ^2 x / cos x = Sin ^2 x / cos x
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#5
mikes gt2:

2(1-cos^2(x))tan(x) = 2(tan(x) - cos(x)sin(x))

how do you work that out?

thanks
0
17 years ago
#6
(Original post by Kurdt Morelo)
mikes gt2:

2(1-cos^2(x))tan(x) = 2(tan(x) - cos(x)sin(x))

how do you work that out?

thanks
2(1-cos^2(x))tan(x) = 2( tan(x) - cos^2(x)tan(x) )

Now, cos^2(x)tan(x) = cos^2(x)[ sin(x)/cos(x) ] = cos(x)sin(x). So,

2( tan(x) - cos^2(x)tan(x) ) = 2( tan(x) - cos(x)sin(x) )

I hope this helps.
0
17 years ago
#7

LHS:
2tan2x-sin2x
= 2( sinx/cosx ) - 2sinxcosx
= [2sinx-2sinxcos(square)x]/cosx
= 2sinx[1-cos(square)x]/cosx
= 2sin(square)xtanx
= RHS
0
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