Prove cosine of sum can be expressed as a linear combination of sine and cosine

Given $A, B \in \mathbb{R}$ , prove that

$C \cos (x+ \alpha) = A \sin x + B \cos x$

for $C, \alpha \in \mathbb{R}$ with $C \geq 0$ .

Further, determine $C, \alpha$ if $A = B = 1$ .

Proof. From Exercise #9 of this section, we know for $A, B \in \mathbb{R}$ , there exist $D, \beta \in \mathbb{R}$ such that

$\begin{align*} && D \sin (x + \beta) &= A \sin x + B \cos x \\ \implies && D \sin \left( \frac{\pi}{2} + x + \left( \beta - \frac{\pi}{2} \right) \right) &= A \sin x + B \cos x \\ \implies && D \cos \left( X + \left( \beta - \frac{\pi}{2} \right)\right) &= A \sin x + B \cos x \end{align*}$

Thus,

$C = D, \quad \alpha = \beta - \frac{\pi}{2} .$

Hence, $C$ and $\alpha$ exist such that $C \cos (x + \alpha) = A \sin x + B \cos x. \qquad \blacksquare$

For the computation, if $A = B = 1$ , then $C = \sqrt{2}$ and $\alpha = \beta - \frac{\pi}{2}$ where $\sin \beta = \cos \beta = \frac{\sqrt{2}}{2} \implies$ \beta = \frac{\pi}{4}$. Thus,

$C = \sqrt{2}, \qquad \alpha = - \frac{\pi}{4}.$

Stumbling Robot

Prove cosine of sum can be expressed as a linear combination of sine and cosine

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply