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burhan101 Group Title

Help me with this trig identity pleasee (

  • 2 years ago
  • 2 years ago

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  1. aceace Group Title
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    Question please

    • 2 years ago
  2. burhan101 Group Title
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    |dw:1353194074865:dw|

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

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

    • 2 years ago
  5. phi Group Title
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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?

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

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

    • 2 years ago
  8. irkiz Group Title
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    combine the LHS into 1 fraction

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

    • 2 years ago
  10. Chlorophyll Group Title
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    Common denom: sin²x -1

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

    • 2 years ago
  12. Praja_01 Group Title
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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

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

    • 2 years ago
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