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lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0how about limits? lol

Traxter
 2 years ago
Best ResponseYou've already chosen the best response.0@seidi.yamauti are you aware of what the graphs of cosx and sinx look like?

Traxter
 2 years ago
Best ResponseYou've already chosen the best response.0@abayomi12 he said without using derivatives.

abayomi12
 2 years ago
Best ResponseYou've already chosen the best response.0I'd simplify this first: f(x) = 2 * sin(x) * cos(x) f(x) = sin(2x)

matricked
 2 years ago
Best ResponseYou've already chosen the best response.0max 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)

mukushla
 2 years ago
Best 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\[\sqrt{a^2+b^2} (\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x)=\sqrt{a^2+b^2} \cos(x+\beta)\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0in other words, use the identity sin(x+y) = sin(x)cos(y)+cos(x)sin(y)

abayomi12
 2 years ago
Best ResponseYou've already chosen the best response.0maximum = sqrt(3^2 + 2^2) = sqrt(13)

CIDACPPRADYUMAN
 2 years ago
Best ResponseYou've already chosen the best response.0अगर आप चाहें तो वोल्फ्रम का उपयोग कर सकते हैं.

CIDACPPRADYUMAN
 2 years ago
Best ResponseYou've already chosen the best response.0@mukushla आप तो मास्टर हो, मास्टर!

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0ye cid walay yahan kya kar rahay hain :P

CIDInspectorDAYA
 2 years ago
Best ResponseYou've already chosen the best response.0@sami21 Kya chal raha hai?

CIDACPPRADYUMAN
 2 years ago
Best ResponseYou've already chosen the best response.0@sami21 chutti pe hain hum sab

CIDACPPRADYUMAN
 2 years ago
Best ResponseYou've already chosen the best response.0humari police

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0btw just to let you know guys irreverent answers are considered as SPAM .

mboorstin
 2 years ago
Best ResponseYou've already chosen the best response.0You 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.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1also \(a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\) for suitable \(\theta\) as i recall you can get this from "addition angle" formula

mboorstin
 2 years ago
Best 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).

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1use \( instead of \[

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you 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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