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

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1. Babynini

@freckles haha sorry me again..with more trig. I've gotten [(2tanx)/(1-tan^2x)][cos^2x]

2. Babynini

Ah no, that's wrong. Let me try again xD

3. Babynini

1) (2tanx)/(sec^2x)

4. Babynini

Gasp!! 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

5. Babynini

@ganeshie8 is this correct? :)

6. Babynini

@jim_thompson5910

7. jim_thompson5910

then you'd turn 2*sin(x)*cos(x) into sin(2x)

8. jim_thompson5910

make sure to keep the right hand side the same the whole time

9. UsukiDoll

after you reach 2sinxcosx use the identity sin2x = 2sinxcosx and you'll have sin2x = sin 2x

10. UsukiDoll

wait... 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

11. Babynini

Right right, I just wanted to make sure the process was correct :)

12. Babynini

um..the second one hahaa sorry yeah typo.

13. UsukiDoll

$\frac{ 2tanx }{ 1+\tan^2x } = \sin2x$ $\frac{ 2tanx }{ \sec^2x } =\sin 2x$ $\frac{ 2\frac{sinx}{cosx} }{ \frac{1}{\cos^2x}} = sin2x$ $2\frac{sinx}{cosx} \times \frac{\cos^2x}{1} = \sin 2x$ the cos x cancels out and 2/1 = 2 $2sinxcosx = 2sinx$ using identity 2sinxcosx = sin2x $\sin2x = \sin2x$

14. UsukiDoll

your attempt is correct...just have to remember that there are A LOT of trig. identities.

15. Babynini

Right...So I had it all right :) I just didn't do the last step to make it into sin2x.

16. Babynini

thanks so much!!

17. freckles

very great work @Babynini

18. Babynini

thank you :)

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