anonymous
  • anonymous
What is the maximun of f(x)=3cosx+2sinx ? Whithout using derivatives
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lgbasallote
  • lgbasallote
how about limits? lol
anonymous
  • anonymous
@seidi.yamauti are you aware of what the graphs of cosx and sinx look like?
anonymous
  • anonymous
@abayomi12 he said without using derivatives.

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anonymous
  • anonymous
ok
anonymous
  • anonymous
I'd simplify this first: f(x) = 2 * sin(x) * cos(x) f(x) = sin(2x)
anonymous
  • anonymous
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)
anonymous
  • anonymous
\[a\cos x+b\sin x=\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)\] 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\[\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)=\sqrt{a^2+b^2} \cos(x+\beta)\]
shubhamsrg
  • shubhamsrg
cool.. B|
phi
  • phi
in other words, use the identity sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
anonymous
  • anonymous
maximum = sqrt(3^2 + 2^2) = sqrt(13)
anonymous
  • anonymous
अगर आप चाहें तो वोल्फ्रम का उपयोग कर सकते हैं.
anonymous
  • anonymous
बात समझ आई?
anonymous
  • anonymous
@mukushla आप तो मास्टर हो, मास्टर!
anonymous
  • anonymous
man this is hindi !
anonymous
  • anonymous
ye cid walay yahan kya kar rahay hain :P
anonymous
  • anonymous
@sami-21 Kya chal raha hai?
anonymous
  • anonymous
@sami-21 chutti pe hain hum sab
anonymous
  • anonymous
humari police
anonymous
  • anonymous
btw just to let you know guys irreverent answers are considered as SPAM .
anonymous
  • anonymous
lol
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
@satellite73, how are you inserting equations without linebreaks? $ signs don't seems to work here (as they would in LaTeX).
anonymous
  • anonymous
use \( instead of \[
anonymous
  • anonymous
Thank you.
anonymous
  • anonymous
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

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