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I´m working on a little paper, and I want to know if it´s possible in any way to solve this:

$$A=B\cdot \cos(x)+C\cdot \sin(x)$$

$A$, $B$ and $C$ are known. I need a way to get the $x$ without using a computer. If not, how do I make it to work on excel? I've tried SOLVER, but it doesn't work...

Thanks in advance

David K
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Paolo
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  • Is "sen" supposed to be $\sin$? See math notation guide. Instead of Excel, I suggest any computer algebra system such as Wolfram Alpha or Sage or... –  Jul 02 '15 at 00:09
  • Is 'sen' sine or secant? (in either case, the answer is yes, however) –  Jul 02 '15 at 00:10
  • The abbreviation sen stands for sine in Spanish. Please see https://es.wikipedia.org/wiki/Seno_(trigonometr%C3%ADa) – Keith Jul 02 '15 at 00:13

2 Answers2

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Here is the outline of an answer.

Divide both sides by $\sqrt{B^2 + C^2}$, so that it becomes $$A' = B' \cos X + C' \sin X,$$ where $B'^2 + C'^2 = 1$. Since $(C',B')$ is a point of the unit circle, there is some angle $\alpha$ for which $\cos \alpha = C'$ and $\sin \alpha = B'$. Now rewrite your equation as $$A' = \sin \alpha \cos X + \cos \alpha \sin X = \sin (X + \alpha).$$ Now your equation is easy to solve.

(An alternative is to write $t = \tan X/2$ and express the right hand side in terms of $t$.)

Keith
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  • One way to find such an angle $\alpha$ is to take $\alpha = \arctan\frac{B'}{C'}.$ – David K Jul 02 '15 at 00:32
  • @DavidK That doesn't work if $C' < 0$. Then you have to add $180^{\circ}$. Another special case is $C' = 0$. – Keith Jul 02 '15 at 00:35
  • Good point, really you should use $\text{atan2}(B',C')$ (two-parameter arc tangent with range $-\pi$ to $\pi$) if $C'\neq 0$. Of course if $C'=0$ none of these tricks are necessary: just take $x=\pm\arccos(A'/B').$ – David K Jul 02 '15 at 00:39
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You can write $A=B\cos (X)+ C\sqrt{1-\cos^2 (X)}$, $(A-B\cos(X))^2=C^2(1-\cos^2(X))$, which is quadratic in $\cos (X)$ As we have squared, we may have introduced a spurious solution.

Ross Millikan
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