Infimum of $\alpha\cos2\theta+\beta\sin2\theta+t_1\cos\theta+t_2\sin\theta$, if $\beta >0$ and $t_1^2+t_2^2=4$

Question

For given $\alpha\in \mathbb{R}, \beta\in \mathbb{R^+}$, and $t_1^2+t_2^2=4$, we define the following function on $[0,2\pi]$: $$\varphi(\theta)=\alpha\cos2\theta+\beta\sin2\theta+t_1\cos\theta+t_2\sin\theta$$ is there a way to determine the infimum of the function $\varphi$ on $[0,2\pi]$?

Answer 1

You can find the infimum (actually the minimum) as follows using:

• Cauchy-Schwarz: $t_1^2+t_2^2 = 4\Rightarrow |t_1\cos\theta+t_2\sin\theta|\leq \sqrt{t_1^2+t_2^2}\cdot \sqrt{\cos^2 \theta + \sin^2\theta}=2$ with $t_1\cos\theta+t_2\sin\theta=-2$ for $\binom{t_1}{t_2}= -2\binom{\cos \theta}{\sin \theta}$.
• $\alpha\cos2\theta+\beta\sin2\theta = \sqrt{\alpha^2+\beta^2}\sin(2\theta + \phi)$ with a phase shift $\phi= \arctan\frac{\alpha}{\beta}$ (since $\beta >0$).

Putting this together you get

• $\min_{\theta \in {[0,2\pi]}}\alpha\cos2\theta+\beta\sin2\theta = -\sqrt{\alpha^2+\beta^2}$ which is attained for $\theta_0 = \frac{1}{2}\left( \frac{3}{2}\pi - \phi\right) \in [0,2\pi]$
• Applying Cauchy-Schwarz and choosing $\binom{t_1}{t_2}= -2\binom{\cos \theta_0}{\sin \theta_0}$ you get $$\min_{\theta \in {[0,2\pi]}} \varphi(\theta)=-\sqrt{\alpha^2+\beta^2}-2$$