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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,

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

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