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Is there any way to solve the following sum of trigonometric functions for theta without using a solver? $$25\sin(\theta)-1.5\cos(\theta)=20$$

JRN
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mjgpy3
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3 Answers3

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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.

Ayman Hourieh
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Michael Hardy
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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.

Eric Angle
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  • I think OP was looking for a solution without using a solver – Daryl Sep 09 '12 at 02:41
  • Eric, but the closed form is readily available. Besides, it shows the multitude of solutions. This answer might attract down-votes, I am afraid. – Sasha Sep 09 '12 at 02:55
  • Then 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. – Eric Angle Sep 09 '12 at 14:15
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The solution to $A\sin(t) + B\cos(t) = C$ is $t = 2\left(\pi n + \arctan{(\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C})}\right)$

Euler....IS_ALIVE
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    Nice. But it'd be more helpful for OP if you show some derivation steps, or a name of a reference. –  Sep 09 '12 at 03:15