A trigonometry problem by Richard Bao

But it is a Calculus problem.

Let $f(x) = 8\sin {x} + 6\cos{x} - 3\quad \Rightarrow \dfrac {d}{dx} f(x) = 8 \cos{x} - 6 \sin{x}$

Equating $\dfrac {d}{dx} f(x) = 0\quad \Rightarrow 8\cos {x} = 6 \sin{x} \quad \Rightarrow \tan{x} = \frac {4}{3}$ $\quad \Rightarrow \sin{x} = \frac {4}{5} \space$ and $\space \cos{x} = \frac {3}{5}$ . Notice that $\frac {d}{dx} f(x) < 0$ , when $\tan{x} = \frac {4}{3}$ .

Therefore, the maximum value of $f(x) = 8(\frac {4}{5}) + 6(\frac {3}{5} ) - 3 = \frac {32+18} {5} - 3 = \boxed{7}$ .

The answer is seven:

the maximum value of 8sin(x) + 6cos(x) is sqrt (8^2 + 6^2) This is the pythagorean theorem imagine 8sin(x) and 6cos(x) as the legs of a right triangle

It can be solved using trigonometry by making 8sin(x)+6cos(x) a single trigonometric function. By using Rsin(x+a)=Rcos(a)sin(x)+Rsin(a)cos(x) and using sin^2(x)+ cos^2(x)=1, so 8sin(x)+6cos(x)=10sin(x+(arctan(3/4)). The maximum value of sin is 1 so the maximum value of 10sin(x+(arctan(3/4)) is 10, therefore maximum value 8sin(x) + 6cos(x) - 3 is 7