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Prove any linear combination of sine and cosine can be expressed solely in terms of sine

With reference to the previous exercise, prove that if A, B \in \mathbb{R} then there exist C, \alpha \in \mathbb{R} with C \geq 0 such that

    \[ A \sin x + B \cos x = C \sin (x + \alpha). \]


Proof. Let

    \[ C = \sqrt{A^2 + B^2} \]

Then, C \in \mathbb{R} and C \geq 0. Further, we have,

    \[ \left| \frac{A}{C} \right| \leq 1 \]

since

    \[ |A| \leq \left| \sqrt{A^2+B^2} \right| \qquad \text{since } B^2 \geq 0. \]

Hence, we know there exists \alpha \in \mathbb{R} such that

    \[ \cos \alpha = \frac{A}{C}. \]

But, if \cos \alpha = \frac{A}{C}, then

    \begin{align*}  \cos^2 \alpha = \frac{A^2}{C^2} && \implies && 1 - \sin^2 \alpha &= \frac{A^2}{A^2+B^2} \\  && \implies && \sin^2 \alpha &= 1 - \frac{A^2}{A^2+B^2} = \frac{B^2}{A^2+B^2} \\ && \implies && \sin \alpha &= \frac{B}{C}. \end{align*}

Thus,

    \begin{align*}    C \sin (x+\alpha) &= C \sin x \cos \alpha + C \sin \alpha \cos x \\  &= (A^2+B^2)^{1/2} \frac{A}{(A^2+B^2)^{1/2}} \sin x + (A^2+B^2)^{1/2} \frac{B}{(A^2+B^2)^{1/2}} \cos x \\  &= A \sin x + B \cos x. \qquad \blacksquare \end{align*}

2 comments

    • Awamoki says:

      The codomain of sin(x) is [-1,1] cause the Pythagorean identity sin^2(theta)+cos^2(theta)=1 and so his solution is legit.

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