I need to convert $b\cos(ax)+c\sin(ax)$ into something like $d\cos(ax-y)$. I thought I could use the identity $\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$, however that's without considering my constants b and c.
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1Perhaps consider what $\sqrt{b^2+c^2}$ represents – Henry Jan 15 '15 at 15:09
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2My answer to this question can help you. – Workaholic Jan 15 '15 at 15:10
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You can use the method of subsidiary angle.
Let $$b = d\cos\phi,$$ $$c=d\sin\phi.$$ In other words, $$d=\sqrt{b^2+c^2},$$ $$\phi=\tan^{-1}\frac{c}{b}.$$ Then $$b\cos(ax)+c\sin(ax)=d\cos\phi\cos(ax)+d\sin\phi\sin(ax)=d\cos(ax-\phi).$$
velut luna
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$$\sqrt{b^2+c^2}\left( \cos \phi \cdot \cos (ax)+\sin \phi \cdot \sin (ax) \right)$$ $$\sqrt{b^2+c^2}\cdot \cos \left( \phi -ax \right)$$ where $\phi=\arccos \dfrac{b}{\sqrt{b^2+c^2}}$
Mihail
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