Maximizing a Trigonometric Expression: asin(x)+bcos(x).
I came across a maximization problem involving trigonometric expressions that I thought was interesting. It might be quite standard, but I have never seen it before. The problem is to maximize the expression \[a \sin x + b \cos x\] where \(a, b \in \R\) are fixed constants and \(x \in \R\) is any real number. Without loss or generality, assume that \(a\neq 0\) and \(b\neq 0\).
Proof and Result
We begin by trying to combine the two terms using trigonometry. Notice that \[r\sin(x+\varphi) = r\cos\varphi\sin x + r\sin\varphi\cos x.\] We can then define \[a = r\cos\varphi, b = r\sin\varphi,\] to obtain that \[a\sin x + b\cos x = r\sin(x+\varphi) \leq r.\] It is clear that the maximum value of the right-hand side occurs when \(\sin(x+\varphi)\) is equal to one and the RHS1 is equal to \(r\). Now it remains for us to find the value of \(r\). Dividing our definitions for \(a\) and \(b\), we see that \[\frac{b}{a} = \frac{r\cos\varphi}{r\sin\varphi} = \tan\varphi,\] and hence, \[\varphi = \arctan\frac{b}{a}.\] We can then plug this in to solve for \(r\). Note that \[\cos\varphi = \cos\arctan\frac{b}{a} = \frac{a}{\sqrt{a^2+b^2}}.\] This result could be obtained by construcing a reference right triangle with opposite side \(b\), adjacent side \(a\), and hence hypotenuse \(\sqrt{a^2+b^2}\). Finally, \[r = \frac{a}{\cos\varphi} = \frac{a}{\frac{a}{\sqrt{a^2+b^2}}} = \sqrt{a^2+b^2},\] and we have found the maximum value of \(a\sin x + b\cos x\). Similarly, since \(\sin(x+\varphi) \geq -1\), we also have that \(a\sin x + b\cos x \geq -r = -\sqrt{a^2 + b^2}\). Therefore, \[\max_{x \in \R}(a\sin x + b\cos x) = \sqrt{a^2 + b^2}, \; \min_{x \in \R}(a\sin x + b\cos x) = -\sqrt{a^2 + b^2}.\]
An Application
This application was inspired by a few comment suggestions under this YouTube Video, The problem presented in the video is as follows. Given that \(x^2 + y^2 = 4\), maximize \(3x+4y\). We immediately notice the potential benefit of trigonometric substitution because the given statement is in the form of the pythagorean identity. Setting \(x = 2\cos\theta\) and \(y = 2\sin\theta\), the given equation becomes \[4\cos^2 \theta + 4\sin^2 \theta = 4,\] which is automatically satisfied by any \(\theta\in\R\). Then, our expression to maximize becomes \[3(2\cos\theta) + 4(2\sin\theta) = 6\cos\theta + 8\sin\theta.\] Applying our result, we see that \[\max_{x^2 + y^2 = 4}(3x+4y) = \max_{\theta \in \R}(8\sin\theta + 6\cos\theta) = \sqrt{8^2 + 6^2} = 10.\] And similarly, its minimum is \(-10\).
We could also visualize this problem in \(\R^3\) (3-space). We are dealing with a ring characterized by the intersection of the cylinder \(x^2 + y^2 = 4\) and the plane \(z = 3x+4y\). In the substitution step, we have effectively parametrized this ring. Taking this one step further, we find that the ring of interest could be fully parametrized as \[\vec{r}(\theta) = \langle 2\cos\theta, 2\sin\theta, 8\sin\theta + 6\cos\theta \rangle,\] where \(\theta\in[0,2\pi)\). Our desired maximum is now the maximum value that our ring achieves in the \(z\)-axis. We could also opt to quickly verify our result using calculus.2