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Babynini

  • one year ago

Prove the identity (2tanx)/(x+tan^2x) = sin2x

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

  2. Babynini
    • one year ago
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    Ah no, that's wrong. Let me try again xD

  3. Babynini
    • one year ago
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    1) (2tanx)/(sec^2x)

  4. Babynini
    • one year ago
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    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
    • one year ago
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    @ganeshie8 is this correct? :)

  6. Babynini
    • one year ago
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    @jim_thompson5910

  7. jim_thompson5910
    • one year ago
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    then you'd turn 2*sin(x)*cos(x) into sin(2x)

  8. jim_thompson5910
    • one year ago
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    make sure to keep the right hand side the same the whole time

  9. UsukiDoll
    • one year ago
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    after you reach 2sinxcosx use the identity sin2x = 2sinxcosx and you'll have sin2x = sin 2x

  10. UsukiDoll
    • one year ago
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    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
    • one year ago
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    Right right, I just wanted to make sure the process was correct :)

  12. Babynini
    • one year ago
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    um..the second one hahaa sorry yeah typo.

  13. UsukiDoll
    • one year ago
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    \[\frac{ 2tanx }{ 1+\tan^2x } = \sin2x \] \[\frac{ 2tanx }{ \sec^2x } =\sin 2x \] \[\frac{ 2\frac{sinx}{cosx} }{ \frac{1}{\cos^2x}} = sin2x \] \[2\frac{sinx}{cosx} \times \frac{\cos^2x}{1} = \sin 2x \] the cos x cancels out and 2/1 = 2 \[2sinxcosx = 2sinx \] using identity 2sinxcosx = sin2x \[\sin2x = \sin2x \]

  14. UsukiDoll
    • one year ago
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    your attempt is correct...just have to remember that there are A LOT of trig. identities.

  15. Babynini
    • one year ago
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    Right...So I had it all right :) I just didn't do the last step to make it into sin2x.

  16. Babynini
    • one year ago
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    thanks so much!!

  17. freckles
    • one year ago
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    very great work @Babynini

  18. Babynini
    • one year ago
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    thank you :)

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