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DLS

  • 2 years ago

I'm very confused about the range of this..

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  1. DLS
    • 2 years ago
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    \[\LARGE y=3|sinx|-2|cosx|\]

  2. DLS
    • 2 years ago
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    I know \[ \LARGE 1=<|sinx|+|cosx|<= \sqrt{2}\]

  3. RadEn
    • 2 years ago
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    y = 3|sinx| - 2|cosx| similar with y = 3|sinx| - 2|sin(90-x)| y_minimum when x = 0, it is y = 3(0) - 2(1) = -2 y_max when x = 90, it is y = 3(1) - 2(0) = 3 so, the range for y is -2 <= y <= 3

  4. DLS
    • 2 years ago
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    why did you convert cosx to sin(90-x) any specific reason?

  5. DLS
    • 2 years ago
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    @RadEn

  6. RadEn
    • 2 years ago
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    that is an identity in trigono :) cos(x) = sin(90-x)

  7. DLS
    • 2 years ago
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    i know that i want to know why did you think of that?

  8. DLS
    • 2 years ago
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    in this question i mean any specific reason?

  9. RadEn
    • 2 years ago
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    i trying to make y in sin terms, so that easier to get the output values maybe :)

  10. DLS
    • 2 years ago
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    hmm

  11. DLS
    • 2 years ago
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    does anyone have any other methods ?just curious..thanks @RadEn though!

  12. RadEn
    • 2 years ago
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    u can make y in cosine terms too y = 3|cos(90-x)| - 2|cosx| you're welcome :)

  13. Shadowys
    • 2 years ago
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    the cosine graph can be though as the sine graph offset by 90 degrees phase. so basically RadEn's method is to simplify the graph. Of course, \(y=3|\sin x| - 2|\cos x|\) y is maximum when the y=3|sin x| is max and y=2|cos x| is min. note that there cannot be negative values. for min y ,it is vice versa. the max value of 3sin x is 3(sinx =1), while the min value of 2 cosx is 0. so that gives 3 the min value 3sin x is 0(sinx =0), while the min value of 2 cosx is 2. so that gives -2 so, -2<=y<=3

  14. DLS
    • 2 years ago
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    thanks!

  15. Shadowys
    • 2 years ago
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    you're welcome :)

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