Write $5 \sin \theta + 12 \cos \theta$ as a single cosine with phase displacement.
I don't know how to start this one. If somebody could give me the formula or a sample that would be amazing!
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JRN
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JustheBeats
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Here's a post I wrote that might help you; it solves the general case. – Milo Brandt May 22 '15 at 00:57
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And Here's a question I asked that solves a related general general case :) – MT_ May 22 '15 at 00:58
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Also, where did you encounter this problem? If you don't know where to start, that context would be useful. – MT_ May 22 '15 at 01:01
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It's the phase displacement. That's what is getting to me. – JustheBeats May 22 '15 at 01:03
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3Weren't the coefficients $5$ and $12$ (instead of $20$ and $17$) like an hour ago? – Ken May 22 '15 at 03:16
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We have $$5\sin(t) + 12\cos(t) = 13 \left(\dfrac5{13}\sin(t)+\dfrac{12}{13}\cos(t)\right)$$ where $13=\sqrt{5^2+12^2}$. Setting $\cos(a) = \dfrac{12}{13}$ and $\sin(a) = \dfrac5{13}$, we obtain $$5\sin(t) + 12\cos(t) = 13 \left(\sin(a)\sin(t)+\cos(a)\cos(t)\right) = 13\cos(t-a)$$ where $a$ is such that $\cos(a) = \dfrac{12}{13}$ and $\sin(a) = \dfrac5{13}$.
Adhvaitha
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@JustheBeats : The reason it is possible to find a number $a$ for which $\sin a=1/13$ and $\cos a=12/13$ is that $(5/13)^2+(12/13)^2=1$, and that works because $\sqrt{5^2+12^2}=13$. That's where $13$ came from. – Michael Hardy May 22 '15 at 03:11