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kartel2012
Group Title
express 3cos(theta)+4sin(theta) in the form r sin(theta+alpha).
 one year ago
 one year ago
kartel2012 Group Title
express 3cos(theta)+4sin(theta) in the form r sin(theta+alpha).
 one year ago
 one year ago

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darthjavier Group TitleBest ResponseYou've already chosen the best response.0
This is a theorem: \[\mathrm{\textrm{If } a,\,b \in \mathbb{R}\textrm{, then:}\\a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\\\textrm{such that: }\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\quad\wedge\quad\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}}\]
 one year ago

darthjavier Group TitleBest ResponseYou've already chosen the best response.0
Proof: \[ \mathrm{ a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\\ a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}(\sin(x)\cos(\theta)+\cos(x)\sin(\theta))\\ a\sin(x)+b\cos(x)=\sqrt{a^2+b^2} (\sin(x)\frac{a}{\sqrt{a^2+b^2}}+\cos(x)\frac{b}{\sqrt{a^2+b^2}})\\ a\sin(x)+b\cos(x)=a\sin(x)+b\cos(x) }\]
 one year ago

darthjavier Group TitleBest ResponseYou've already chosen the best response.0
In your question: \[\mathrm{3\cos(\theta)+4\sin(\theta)=5\sin(\theta+\alpha)}\] Where: \[\mathrm{\sin(\alpha)=\frac{3}{5}\quad\wedge\quad\cos(\alpha)=\frac{4}{5}}\]
 one year ago
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