Is there any way to solve the following sum of trigonometric functions for theta without using a solver? $25\sin(\theta)-1.5\cos(\theta)=20$
Solve sum for theta
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trigonometry
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8Relevant? [In the equation $x\cos(\theta) + y\sin(\theta) = z$ how do I solve in terms of $\theta$?](http://math.stackexchange.com/questions/33150/in-the-equation-x-cos-theta-y-sin-theta-z-how-do-i-solve-in-terms-of?rq=1) – 2012-09-09
3 Answers
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$ 25\sin\theta-1.5\cos\theta = \sqrt{25^2+1.5^2}\left( \frac{25}{\sqrt{25^2+1.5^2}}\sin\theta - \frac{1.5}{\sqrt{25^2+1.5^2}}\cos\theta \right) $ $ = \sqrt{25^2+1.5^2}(\cos\varphi\sin\theta-\sin\varphi\cos\theta) = \sqrt{25^2+1.5^2} \sin(\varphi-\theta). $ So you want $ \sin(\varphi-\theta)=\frac{20}{\sqrt{25^2+1.5^2}}. $ Take arcsines.
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0+1. I just fixed the plus in the first set of parentheses. – 2012-09-09
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If you're okay with a numeric solution, define a function $ f\left(\theta\right) = 25 \sin \theta - 1.5 \cos \theta - 20 = 0 $ and use a root finding algorithm.
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0Then the OP should've asked for an analytic solution. To me, a "solver" is something like Mathematica. Besides, I qualified my answer with the phrase "if you're okay with a numeric solution." Finally, it's pretty clear that $f\left(\theta^*+2 \pi n\right) = 0$, where $\theta^*$ is a solution and $n$ is an integer. – 2012-09-09
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The solution to $A\sin(t) + B\cos(t) = C$ is $t = 2\left(\pi n + \arctan{(\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C})}\right)$
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1Nice. But it'd be more helpful for OP if you show some derivation steps, or a name of a reference. – 2012-09-09