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math>philosophy

  • 3 years ago

Prove that the range of \[1 + \sin ^{2} x\] is between 1 and 2, inclusive. How do I show this?

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  1. AravindG
    • 3 years ago
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    sin x maximum vallue is 1

  2. AravindG
    • 3 years ago
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    so max value of exp will be 2

  3. AravindG
    • 3 years ago
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    now minimum value when x=0 sin x=0

  4. AravindG
    • 3 years ago
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    so minimum value =1

  5. AravindG
    • 3 years ago
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    so 1+x^2 has values [1,2]

  6. AravindG
    • 3 years ago
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    gt tht?

  7. AravindG
    • 3 years ago
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    becaus sin x max value is 1

  8. akash_809
    • 3 years ago
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    sinx , minimum value =-1, sin^2x , minimum value =0 ( sin^2x) cant be negative and maximum value of sin^2x=1....put sin^2x=1 for maximum value and put sin^2x=0 for minimum value

  9. akash_809
    • 3 years ago
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    yes maximum value is 2

  10. AravindG
    • 3 years ago
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    so to maximise sin^x+1 ,,,,sin x maximum should be there

  11. AravindG
    • 3 years ago
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    ie 1+1=2

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