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ParthKohli

  • one year ago

I will state the question exactly as it is:

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  1. ParthKohli
    • one year ago
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    If \(x,y \in [0, \pi/2)\):\[\sin^4 x + \cos^4 y + 2 = 4\sin x \cos y\]Then find the value of:\[\sin x + \sin y \]... The answer is given as 2.

  2. anonymous
    • one year ago
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    @Mehek14 @pinkbubbles

  3. ParthKohli
    • one year ago
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    I have two issues with this: one, \(\pi/2\) is not included. If I acknowledge this as a small mistake, even then, if \(\sin x + \sin y = 2\), then \(x = y = \pi/2\). But of course this does not satisfy the condition. Here is what *does* satisfy the condition: \(x = 0;~ y = \pi/2\)

  4. ParthKohli
    • one year ago
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    @ganeshie8

  5. ParthKohli
    • one year ago
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    @ikram002p @dan815

  6. anonymous
    • one year ago
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    @SolomonZelman

  7. ParthKohli
    • one year ago
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    @oldrin.bataku of course.

  8. ParthKohli
    • one year ago
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    Oops, I meant that \(y = 0; x = \pi/2\) satisfies the condition. Moving on...

  9. imqwerty
    • one year ago
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    Pi/2 MUST be included.

  10. imqwerty
    • one year ago
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    I've solved the question thoroughly nd m sure that pi/2 mst be included.

  11. ParthKohli
    • one year ago
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    Even then, the answer couldn't be 2, right?

  12. ParthKohli
    • one year ago
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    Also, how did you solve this question thoroughly if you don't mind explaining?

  13. imqwerty
    • one year ago
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    K wait lemme switch on my laptop :)

  14. ParthKohli
    • one year ago
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    Did you solve this my maxima-minima or ...?

  15. imqwerty
    • one year ago
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    No i jst simplified the expression

  16. ParthKohli
    • one year ago
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    Oh, great.

  17. imqwerty
    • one year ago
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  18. ParthKohli
    • one year ago
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    Yeah, that's what I got too... finally.

  19. ParthKohli
    • one year ago
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    Thank you!

  20. ParthKohli
    • one year ago
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    So the answer isn't 2, it's 1, right?

  21. ParthKohli
    • one year ago
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    The question asked for sin x + sin y...

  22. imqwerty
    • one year ago
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    yes its one :)

  23. ParthKohli
    • one year ago
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    Cool, I was wondering where the mistake was in my work. There wasn't any. I'll report this, thanks!

  24. imqwerty
    • one year ago
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    ur welcome :)

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