Is there a way to find the minimum value of the equation $$y = (4\sin x – 6\cos x)^2 – 3$$ without using calculus or graphing the function?
I wrote $4\sin x – 6\cos x$ in the form $R\sin(x-a)$ where $a$ is a constant. I got $2\sqrt{13}\sin(x-56.31^\circ)$. However, I am unable to find the least value of the given equation but I could find the greatest value as $(2\sqrt{13})^2 – 3$ as the greatest value of sine function is $1$ so a factor of $2\sqrt{13}$ multiplied and then squared would give me 52 then a vertical translation by vector $(0,-3)$ would give the max value as 49. Using a similar approach by knowing the least value of sine function is $-1$, I could not work out the minimum of the original equation.
Best Answer
Since the square of any real number is nonnegative, $$(4\sin x-6\cos x)^2\geq0$$ Add $-3$ to both sides $$f(x)\geq -3$$ So, the minimum value of $f$ is $\boxed{-3}$.
Since your title says a different thing, let me write that out too. As you said, we can write $$a\sin x\pm b\cos x =d\sin (x\pm c)$$ where $d=\sqrt{a^2+b^2}$, from the derivation (I presume you know this as you used it).
Since the sine lies between $-1$ and $1$, $$-1\leq \sin (x-c)\leq 1$$ Multiplying by $d$ (square root is always positive and so is $d$), $$-d\leq d\sin(x-c)\leq d$$ or $$\boxed{-\sqrt{a^2+b^2}\leq a\sin x\pm b\cos x\leq\sqrt{a^2+b^2}}$$
Hope this helps. Ask anything if not clear :)