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Babynini
 one year ago
Prove the identity
(2tanx)/(x+tan^2x) = sin2x
Babynini
 one year ago
Prove the identity (2tanx)/(x+tan^2x) = sin2x

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.2@freckles haha sorry me again..with more trig. I've gotten [(2tanx)/(1tan^2x)][cos^2x]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Ah no, that's wrong. Let me try again xD

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Gasp!! I think I got it :) \[\frac{ 2tanx }{ 1+\tan^2x }\] \[\frac{ 2*\frac{ sinx }{ cosx } }{ \frac{ 1 }{ \cos^2x } }\] \[2*\frac{ sinx }{ cosx }*\frac{ \cos^2x }{ 1 }\] = simplify and cross out a few things 2sinxcosx

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 is this correct? :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0then you'd turn 2*sin(x)*cos(x) into sin(2x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0make sure to keep the right hand side the same the whole time

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2after you reach 2sinxcosx use the identity sin2x = 2sinxcosx and you'll have sin2x = sin 2x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2wait... is there a typo somewhere? you had \[\frac{2tanx}{x+\tan^2x}\] in your original question and then all of a sudden I see \[\frac{2tanx}{1+\tan^2x}\] I am crossing fingers here because if that x+tan^2x turns out to be a typo then everything will work out

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Right right, I just wanted to make sure the process was correct :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2um..the second one hahaa sorry yeah typo.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ 2tanx }{ 1+\tan^2x } = \sin2x \] \[\frac{ 2tanx }{ \sec^2x } =\sin 2x \] \[\frac{ 2\frac{sinx}{cosx} }{ \frac{1}{\cos^2x}} = sin2x \] \[2\frac{sinx}{cosx} \times \frac{\cos^2x}{1} = \sin 2x \] the cos x cancels out and 2/1 = 2 \[2sinxcosx = 2sinx \] using identity 2sinxcosx = sin2x \[\sin2x = \sin2x \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2your attempt is correct...just have to remember that there are A LOT of trig. identities.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.2Right...So I had it all right :) I just didn't do the last step to make it into sin2x.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0very great work @Babynini
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