MHB Linear combination of sine and cosine function

[anemone]

Hi MHB! I recently came across a problem and I was thinking most likely I was missing something very obvious because I couldn't make sense of what was being asked, and I so wish to know what exactly that I failed to relate.

Question:
Find the minimum of $6\sin x+8\cos x+5$. Hence, find the minimum of $(6\sin x+8\cos x)^2+5,\,(6\sin x+8\cos x)^3+5$ and $(6\sin x+8\cos x)^4+5$.

It is important to stress that students are expected to solve it via trigonometry route but not other methods.

I would feel the "hence" implies that the first part of the question greatly help to reach to the answers for the subsequent parts of the problem, but not that I could see it...therefore, any insight would be helpful and thanks in advanced for the help!

 

[HOI]

If f(x) has minimum m at x= x0 then f(x)+ 5 has minimum m+ 5 at x0, f(x)^2+ 5 has minimum m^2+ 5 at x= x0, and, generally, f(x)^n+ 5 has minimum m^n+ 5 at x= x-x0.

 

[Opalg]

If f(x) has minimum m at x= x0 then[/color] f(x)+ 5 has minimum m+ 5 at x0, f(x)^2+ 5 has minimum m^2+ 5 at x= x0[/color], and, generally, f(x)^n+ 5 has minimum m^n+ 5 at x= x-x0.

That is not true. For example, the function $f(x) = \sin x$ has minimum $m = -1$ at $x_0 = \frac{3\pi}2$. But $f(x)^2 + 5$ does not have minimum $m^2+5 = 6$ at $x = x_0$. Instead, it has minimum $5$ at $x = 0$.

 

[anemone]

Argh!(Headbang) I couldn't believe how I overlooked something so trivially simple in order to deduce the minimum of the other functions! Once we rewrite $6\sin x+8\cos x+5=10\sin \left(x+\tan^{-1}\dfrac{4}{3}\right)+5$, we know the minimum can be attained at $-5$, and so the minimum values of

$(6\sin x+8\cos x)^2+5=\left(10\sin \left(x+\tan^{-1}\dfrac{4}{3}\right)\right)^2+5$ is $0+5=5$,

$(6\sin x+8\cos x)^3+5=\left(10\sin \left(x+\tan^{-1}\dfrac{4}{3}\right)\right)^3+5$ is $(-10)^3+5=-995$,

$(6\sin x+8\cos x)^4+5=\left(10\sin \left(x+\tan^{-1}\dfrac{4}{3}\right)\right)^4+5$ is $0+5=5$.

Sorry for asking something so simple!