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seidi.yamauti

What is the maximun of f(x)=3cosx+2sinx ? Whithout using derivatives

  • one year ago
  • one year ago

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  1. lgbasallote
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    how about limits? lol

    • one year ago
  2. Traxter
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    @seidi.yamauti are you aware of what the graphs of cosx and sinx look like?

    • one year ago
  3. Traxter
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    @abayomi12 he said without using derivatives.

    • one year ago
  4. abayomi12
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    ok

    • one year ago
  5. abayomi12
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    I'd simplify this first: f(x) = 2 * sin(x) * cos(x) f(x) = sin(2x)

    • one year ago
  6. matricked
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    max value of acosx+bsinx is sqrt(a^2+b^2) and min is -sqrt(a^2+b^2) thus here max value is sqrt(3^2+2^2)

    • one year ago
  7. mukushla
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    \[a\cos x+b\sin x=\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)\] using the fact that\[(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1\] u can prove\[\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)=\sqrt{a^2+b^2} \sin(x+\alpha)\]or\[\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)=\sqrt{a^2+b^2} \cos(x+\beta)\]

    • one year ago
  8. shubhamsrg
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    cool.. B|

    • one year ago
  9. phi
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    in other words, use the identity sin(x+y) = sin(x)cos(y)+cos(x)sin(y)

    • one year ago
  10. abayomi12
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    maximum = sqrt(3^2 + 2^2) = sqrt(13)

    • one year ago
  11. CID-ACP-PRADYUMAN
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    अगर आप चाहें तो वोल्फ्रम का उपयोग कर सकते हैं.

    • one year ago
  12. CID-ACP-PRADYUMAN
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    बात समझ आई?

    • one year ago
  13. CID-ACP-PRADYUMAN
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    @mukushla आप तो मास्टर हो, मास्टर!

    • one year ago
  14. mukushla
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    man this is hindi !

    • one year ago
  15. sami-21
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    ye cid walay yahan kya kar rahay hain :P

    • one year ago
  16. CID-Inspector-DAYA
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    @sami-21 Kya chal raha hai?

    • one year ago
  17. CID-ACP-PRADYUMAN
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    @sami-21 chutti pe hain hum sab

    • one year ago
  18. CID-ACP-PRADYUMAN
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    humari police

    • one year ago
  19. sami-21
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    btw just to let you know guys irreverent answers are considered as SPAM .

    • one year ago
  20. CID-ACP-PRADYUMAN
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    lol

    • one year ago
  21. mboorstin
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    You can use a special trig identity: Acos(x) + Bsin(x) = Ccos(x-y), where C=\sqrt{A^2=B^2) and y = arctan(B/A). Hence \[f(x)=3\cos x + 2\sin x = \sqrt{13}\cos\left(x-arctan\tfrac{2}{3}\right)\approx\sqrt{13}\cos\left(x-.588\right)\] But in any case, the maximum for cos is 1, so the maximum for f(x) is the square root of 13.

    • one year ago
  22. satellite73
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    also \(a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\) for suitable \(\theta\) as i recall you can get this from "addition angle" formula

    • one year ago
  23. mboorstin
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    @satellite73, how are you inserting equations without linebreaks? $ signs don't seems to work here (as they would in LaTeX).

    • one year ago
  24. satellite73
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    use \( instead of \[

    • one year ago
  25. mboorstin
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    Thank you.

    • one year ago
  26. satellite73
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    you are right, $ does not work here turns out \( is an alternative in latex if you need to see any code, right click and you can see it. it is good method for copying and pasting as well, so you don't have to rewrite the latex every time

    • one year ago
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