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burhan101

  • 3 years ago

Help me with this trig identity pleasee (

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  1. aceace
    • 3 years ago
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    Question please

  2. burhan101
    • 3 years ago
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    |dw:1353194074865:dw|

  3. aceace
    • 3 years ago
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    simplyfy cos ^2 x... what is that now?

  4. aceace
    • 3 years ago
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    clue sin^2 x + sin^2 x = 1

  5. phi
    • 3 years ago
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    I think ace meant \[ \sin^2 x + \cos^2 x = 1\] I would first find a common denominator. Can you do that?

  6. aceace
    • 3 years ago
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    Yes @phi that is exactly what i meant XD

  7. burhan101
    • 3 years ago
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    @aceace @phi i would get this for the right side : \[\huge =\frac{ 2 }{ 1-\sin^2x }\]

  8. irkiz
    • 3 years ago
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    combine the LHS into 1 fraction

  9. gohangoku58
    • 3 years ago
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    1-sin^2 x = cos^2 x

  10. Chlorophyll
    • 3 years ago
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    Common denom: sin²x -1

  11. Chlorophyll
    • 3 years ago
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    After simplify you'll get the result = 2/ cos²x

  12. Praja_01
    • 3 years ago
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    1/sinx+1=sinx-1/sin^2x-1 =1-sinx/1-sin^2x =1-sinx/cos^2x 1/sinx-1=sinx+1/sin^2x-1 =-1-sinx/1-sin^2x =-1-sinx/cos^2x =(1-sinx-(-1-sinx)/cos^2x =2/cos^x

  13. aceace
    • 3 years ago
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    sorry i was away... but that would be correct XD

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