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DanielaJohana

  • one year ago

use trigonometric identities to simplify: (csc(x)-tan(x))sin(x)cos(x)

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  1. Nnesha
    • one year ago
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    write csc and tan in terms of sin and cos

  2. Nnesha
    • one year ago
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    csc =?? tan= ?? what is reciprocal of tan and csc ?

  3. DanielaJohana
    • one year ago
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    reciprocal of tanx is 1/cotx reciprocal of cscx is 1/sinx ?

  4. DanielaJohana
    • one year ago
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    @Nnesha

  5. Nnesha
    • one year ago
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    yes right and also tan = sin over cos \[\huge\rm [\color{red}{\csc(x)-\tan(x)}]\sin(x)\cos(x)\] \[(\color{Red}{\frac{ 1 }{ \sin(x)} -\frac{ \sin(x) }{ \cos(x) }})\sin(x)\cos(x)\] now solve the parentheses find common denomiantor

  6. Nnesha
    • one year ago
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    can you do that ?? :)

  7. DanielaJohana
    • one year ago
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    @Nnesha So the common denom would be sin(x)cos(x) right?

  8. Nnesha
    • one year ago
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    yep

  9. Nnesha
    • one year ago
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    \[(\color{Red}{\frac{ ??-?? }{ \sin(x)cos(x)} })\sin(x)\cos(x)\] multiply top of first fraction by the bottom of the 2nd fraction multiply numerator of 2nd fraction by denominator of first fraction \[(\color{Red}{\frac{ 1(cos(x))-sin(x)sin(x) }{ \sin(x)cos(x)} })\sin(x)\cos(x)\]

  10. Nnesha
    • one year ago
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    try to simplify tht from there

  11. DanielaJohana
    • one year ago
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    @Nnesha Would I cancel the denominator with the outside?

  12. Nnesha
    • one year ago
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    yes you can

  13. DanielaJohana
    • one year ago
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    @Nnesha so would my final answer be \[\cos(x)-\sin^{2}(x)\]

  14. Nnesha
    • one year ago
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    you can simplify that sin^2 = ? btw do you have options ?

  15. DanielaJohana
    • one year ago
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    @Nnesha no I don't \[\sin ^{2}x = \frac{ 1-\cos2(x) }{ 2 }\]

  16. Nnesha
    • one year ago
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    sin^2x is just equal to 1-cos^2\[\rm cos^2 \theta +\sin^2 \theta =1\] so when you solve for sin^2theat you will get 1-cos^2

  17. Nnesha
    • one year ago
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    but i don't think we should substitute sin^2 just leave it as cos(x)-sin^2(x)

  18. DanielaJohana
    • one year ago
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    oh okay then thanks

  19. Nnesha
    • one year ago
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    my pleasure

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