Just Observed A Cool Result!

I just found a generalised formula to find the range of the functions of the form: f(x)=acosx+bsinx+cf(x)=a\cos x + b\sin x + c

The range of the above function would be: [a2+b2+c,a2+b2+c]\left[-\sqrt{a^2+b^2}+c, \sqrt{a^2+b^2}+c\right].


Here's what I did;

1. Multiply and divide the function by a2+b2\sqrt{a^2 + b^2};

f(x)=a2+b2(acosxa2+b2+bsinxa2+b2)+cf(x)=\sqrt{a^2 + b^2}\left(\dfrac{a\cos x}{\sqrt{a^2 + b^2}} + \dfrac{b\sin x}{\sqrt{a^2 + b^2}}\right)+c

2. Now, say that there exists a right triangle whose sides are a,b,a2+b2a, b, \sqrt{a^2 + b^2}, such that;

sinβ=aa2+b2;cosβ=ba2+b2\sin\beta = \dfrac{a}{\sqrt{a^2 + b^2}};\quad \cos\beta = \dfrac{b}{\sqrt{a^2 + b^2}}

3. Substitute the above values in the function to get:

f(x)=a2+b2(sinβcosx+cosβsinx)+c=a2+b2sin(β+x)+c\begin{aligned} f(x)&=\sqrt{a^2 + b^2}\left(\sin\beta\cos x+ \cos\beta\sin x\right) + c\\ &=\sqrt{a^2 + b^2}\sin(\beta +x)+c\\ \end{aligned}

4. Now let us say that the angle β+x=α\beta + x=\alpha, thus the maximum value of sinα=1\sin\alpha = 1 and the minimum value of sinα=1\sin\alpha = -1, hence;

f(x)max=a2+b2+cf(x)min=a2+b2+c\begin{aligned} f(x)_{\text{max}}&=\sqrt{a^2 + b^2}+c\\ f(x)_{\text{min}}&=-\sqrt{a^2 + b^2}+c\\ \end{aligned}

Note by Sravanth Chebrolu
3 years, 4 months ago

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That is a general method in practice. Anyway finding it on your own is great! !

Abhi Kumbale - 3 years, 4 months ago

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Oh I thought so, thanks!

Sravanth Chebrolu - 3 years, 4 months ago

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I try...we have f(x)=acosx+bsinx+cf(x)=a\cos x + b\sin x + c If we derive we obtain f(x)=asinx+bcosxf'(x)=-a\sin x + b\cos x To find max/min asinx+bcosx=0-a\sin x + b\cos x=0 We find cosx=absinx\cos x=\frac{a}{b}\sin x tanx=ba\tan x=\frac{b}{a} So now we can replace them in f(x) ymax=(a2+b2b)sin(tan1(ba))+cy_{max}=(\frac{a^2+b^2}{b})\sin(\tan^{-1}(\frac{b}{a})) + c

but if we consider a rectangle triangle with sides a, b and c, where alpha is the angle between b and c (and the angle between a and b is 90°) then

sin(tan1(ba))=bc=ba2+b2\sin(\tan^{-1}(\frac{b}{a}))=\frac{b}{c}=\frac{b}{\sqrt{a^2+b^2}}

So now

ymax=a2+b2+cy_{max}=\sqrt{a^2+b^2} + c

We can try to see that it is really the maximum of f(x), so the minimum symmetrically must be

ymin=a2+b2+cy_{min}=-\sqrt{a^2+b^2} + c

Right...?

Matteo Monzali - 3 years, 4 months ago

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Exactly! I almost did the same! Splendid!

I have reduced some steps, thanks for posting it!

Sravanth Chebrolu - 3 years, 4 months ago

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For an elegant proof try using Cauchy Schwarz Inequality.

Sal Gard - 3 years, 4 months ago

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@Matteo Monzali: Since you have posted this, I have given another version of what I did. Please check that out.

Sravanth Chebrolu - 3 years, 4 months ago

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Yes, but I think your method is easier than mine! :)

Matteo Monzali - 3 years, 4 months ago

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