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seidi.yamauti
What is the maximun of f(x)=3cosx+2sinx ? Whithout using derivatives
how about limits? lol
@seidi.yamauti are you aware of what the graphs of cosx and sinx look like?
@abayomi12 he said without using derivatives.
I'd simplify this first: f(x) = 2 * sin(x) * cos(x) f(x) = sin(2x)
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)
\[a\cos x+b\sin x=\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)\] 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\[\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)=\sqrt{a^2+b^2} \cos(x+\beta)\]
in other words, use the identity sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
maximum = sqrt(3^2 + 2^2) = sqrt(13)
अगर आप चाहें तो वोल्फ्रम का उपयोग कर सकते हैं.
@mukushla आप तो मास्टर हो, मास्टर!
ye cid walay yahan kya kar rahay hain :P
@sami-21 Kya chal raha hai?
@sami-21 chutti pe hain hum sab
humari police
btw just to let you know guys irreverent answers are considered as SPAM .
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.
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
@satellite73, how are you inserting equations without linebreaks? $ signs don't seems to work here (as they would in LaTeX).
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