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

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

Here's what I did;

**1.** Multiply and divide the function by \(\sqrt{a^2 + b^2}\);

\[f(x)=\sqrt{a^2 + b^2}\left(\dfrac{a\cos x}{\sqrt{a^2 + b^2}} + \dfrac{b\sin x}{\sqrt{a^2 + b^2}}\right)+c\]

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

\[\sin\beta = \dfrac{a}{\sqrt{a^2 + b^2}};\quad \cos\beta = \dfrac{b}{\sqrt{a^2 + b^2}}\]

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

\[\begin{align} f(x)&=\sqrt{a^2 + b^2}\left(\sin\beta\cos x+ \cos\beta\sin x\right) + c\\ &=\sqrt{a^2 + b^2}\sin(\beta +x)+c\\ \end{align}\]

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

\[\begin{align} f(x)_{\text{max}}&=\sqrt{a^2 + b^2}+c\\ f(x)_{\text{min}}&=-\sqrt{a^2 + b^2}+c\\ \end{align}\]

## Comments

Sort by:

TopNewestThat is a general method in practice. Anyway finding it on your own is great! ! – Abhi Kumbale · 1 year, 3 months ago

Log in to reply

– Sravanth Chebrolu · 1 year, 3 months ago

Oh I thought so, thanks!Log in to reply

I try...we have \[f(x)=a\cos x + b\sin x + c\] If we derive we obtain \[f'(x)=-a\sin x + b\cos x\] To find max/min \[-a\sin x + b\cos x=0\] We find \[\cos x=\frac{a}{b}\sin x\] \[\tan x=\frac{b}{a}\] So now we can replace them in f(x) \[y_{max}=(\frac{a^2+b^2}{b})\sin(\tan^{-1}(\frac{b}{a})) + c\]

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

\[\sin(\tan^{-1}(\frac{b}{a}))=\frac{b}{c}=\frac{b}{\sqrt{a^2+b^2}}\]

So now

\[y_{max}=\sqrt{a^2+b^2} + c\]

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

\[y_{min}=-\sqrt{a^2+b^2} + c\]

Right...? – Matteo Monzali · 1 year, 3 months ago

Log in to reply

@Matteo Monzali: Since you have posted this, I have given another version of what I did. Please check that out. – Sravanth Chebrolu · 1 year, 3 months ago

Log in to reply

– Matteo Monzali · 1 year, 3 months ago

Yes, but I think your method is easier than mine! :)Log in to reply

I have reduced some steps, thanks for posting it! – Sravanth Chebrolu · 1 year, 3 months ago

Log in to reply

– Sal Gard · 1 year, 3 months ago

For an elegant proof try using Cauchy Schwarz Inequality.Log in to reply