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prove tan x cos 2x + tan x + sin 2 x

nlbradshaw0430  Apr 13, 2017
edited by nlbradshaw0430  Apr 13, 2017
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4+0 Answers

 #1
avatar+76959 
+1

I think you're missing something, here......there's nothing to prove....

 

 

cool cool cool

CPhill  Apr 13, 2017
edited by CPhill  Apr 13, 2017
 #2
avatar+90571 
0

Maybe it should just be simplify????

 

 tan x cos 2x + tan x + sin 2 x

 

\(tan x *cos 2x + tan x + sin 2 x\\ =\frac{sinx}{cosx} *(cos^2x-sin^2x)+ \frac{sinx}{cosx} + 2sinxcosx\\ =sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} + 2sinxcosx\\ =3sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} \\ =3sinxcosx- \frac{sinx}{cosx} (sin^2x-1)\\ =3sinxcosx- \frac{sinx}{cosx} (-cos^2x)\\ =3sinxcosx+ (sinxcosx)\\ =4sinxcosx\\ =2sin2x \)

 

Lets see if I could have done this more simply....

 

\( tan x cos 2x + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x) + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x+1) + sin 2 x\\ = tan x (cos^2x+cos^2x) + sin 2 x\\ =\frac{sinx}{cosx} (2cos^2x) + sin 2 x\\ =2sinxcosx + sin 2 x\\ =sin2x + sin 2 x\\ =2sin2x \\\)

 

So it can be seen that 

tan x cos 2x + tan x = sin 2 x

which I guess was meant to be the question in the first place.   (as hectictar and CPhill have already noted.  Thanks guys. :)

Melody  Apr 13, 2017
edited by Melody  Apr 13, 2017
 #3
avatar+76959 
+1

tan x cos 2x + tan x + sin 2 x      ......simplified  ???

 

[sinx/ cosx] [ cos^2x - sin^2x]  + sinx/cosx +  2sinx cosx

 

sinxcosx  - sin^3x/ cosx   +  sinx/cosx + 2sinxcosx

 

3sinxcosx  -  [sinx ( sin^2x - 1)] / cosx

 

3sinxcosx  -  [ -sinx (1 - sin^2x) ]/ cosx

 

3sinxcosx  + [sinx cos^2x]/cosx

 

3sinxcosx + sinxcosx  =

 

4sinxcosx       or       2sin(2x)

 

 

 

cool cool cool

CPhill  Apr 13, 2017
 #4
avatar+76959 
+2

OK...I see now....

 

tan x cos 2x + tan x = sin 2 x

 

tanx (cos 2x + 1)

 

tan x ( 2cos^2x - 1 + 1)

 

tanx ( 2cos^2x)

 

2 * [sinx / cosx] * cos^2x 

 

2sin x cos x    =    sin 2x

 

Thanks to hectictar for bringing this to my attention....!!!!

 

 

cool cool cool

CPhill  Apr 13, 2017
edited by CPhill  Apr 13, 2017

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