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

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lgbasalloteBest ResponseYou've already chosen the best response.0
how about limits? lol
 one year ago

TraxterBest ResponseYou've already chosen the best response.0
@seidi.yamauti are you aware of what the graphs of cosx and sinx look like?
 one year ago

TraxterBest ResponseYou've already chosen the best response.0
@abayomi12 he said without using derivatives.
 one year ago

abayomi12Best ResponseYou've already chosen the best response.0
I'd simplify this first: f(x) = 2 * sin(x) * cos(x) f(x) = sin(2x)
 one year ago

matrickedBest ResponseYou've already chosen the best response.0
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

mukushlaBest ResponseYou've already chosen the best response.7
\[a\cos x+b\sin x=\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)\] 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

phiBest ResponseYou've already chosen the best response.0
in other words, use the identity sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
 one year ago

abayomi12Best ResponseYou've already chosen the best response.0
maximum = sqrt(3^2 + 2^2) = sqrt(13)
 one year ago

CIDACPPRADYUMANBest ResponseYou've already chosen the best response.0
अगर आप चाहें तो वोल्फ्रम का उपयोग कर सकते हैं.
 one year ago

CIDACPPRADYUMANBest ResponseYou've already chosen the best response.0
@mukushla आप तो मास्टर हो, मास्टर!
 one year ago

sami21Best ResponseYou've already chosen the best response.0
ye cid walay yahan kya kar rahay hain :P
 one year ago

CIDInspectorDAYABest ResponseYou've already chosen the best response.0
@sami21 Kya chal raha hai?
 one year ago

CIDACPPRADYUMANBest ResponseYou've already chosen the best response.0
@sami21 chutti pe hain hum sab
 one year ago

CIDACPPRADYUMANBest ResponseYou've already chosen the best response.0
humari police
 one year ago

sami21Best ResponseYou've already chosen the best response.0
btw just to let you know guys irreverent answers are considered as SPAM .
 one year ago

mboorstinBest ResponseYou've already chosen the best response.0
You can use a special trig identity: Acos(x) + Bsin(x) = Ccos(xy), where C=\sqrt{A^2=B^2) and y = arctan(B/A). Hence \[f(x)=3\cos x + 2\sin x = \sqrt{13}\cos\left(xarctan\tfrac{2}{3}\right)\approx\sqrt{13}\cos\left(x.588\right)\] 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

satellite73Best ResponseYou've already chosen the best response.1
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

mboorstinBest ResponseYou've already chosen the best response.0
@satellite73, how are you inserting equations without linebreaks? $ signs don't seems to work here (as they would in LaTeX).
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
use \( instead of \[
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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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