How to solve $A\sin{\theta} + B\cos{\theta} = C$?

Question

I've stumbled upon a equation in the form $$A\sin{\theta} + B\cos{\theta} = C$$ What would be the steps necessary to solving it?

Thank you.

Obs: $A$, $B$ and $C$ are known; the unknown variable is $\theta$. — DaviFN, Sep 28 '20 at 22:54
the best approach is divide everything with $C$, and if you are lucky, there will exists an angle $\alpha$ such that $\cos\alpha = \frac{A}{C}$ and $\sin\alpha = \frac{B}{C}$. So you have $\sin(\alpha+\theta) = 1$ — L F, Sep 28 '20 at 22:57
Well yes, ... but that's a different $C$, possibly, from the one given in the problem. — John Hughes, Sep 28 '20 at 22:59
This question is a duplicate of MSE question 2502976 "Solving $a \sin\theta + b \cos\theta = c$". — Somos, Sep 29 '20 at 12:56

John Hughes · Answer 1 · 2020-09-28T23:27:18.030

1

Typical is to let $u = \arctan(A/B)$ (dealing with the special case of $B = 0$ separately), $r = \sqrt{A^2 + B^2}$. Then \begin{align} A &= r \sin u\\ B &= r \cos u \end{align} (unless I've swapped those two). Now your equation reads $$ r \sin u \sin \theta + r \cos u \cos \theta = C $$ which you rewrite as \begin{align} \cos(u-\theta) &= \frac{C}{r}\\ \theta &= u - \arccos(\frac{C}{r}) \end{align} and you're done.

NB: If $|C| > |r|$, then there is no (real) solution. (Hat-tip to @LuisFelipe for making me add the "real".) If $r$ is zero and $C$ is nonzero, there's also no solution. If $r = C = 0$, then every value of $\theta$ is a solution. That leaves only the special case $B = 0, A \ne 0$ for you to work out.

edited Sep 28 '20 at 23:27

answered Sep 28 '20 at 22:58

John Hughes

93,729

there is always solution in $\mathbb{C}$ :D – L F Sep 28 '20 at 23:01
yes ... and if the OP had wanted that, I'll bet that OP could have solved the problem this way without my help. – John Hughes Sep 28 '20 at 23:03
There's no solution in $\mathbb{C}$ if $A=B=0$ and $C=1$ ;) – Matthew H. Sep 28 '20 at 23:17

score 0 · Answer 2 · answered Sep 28 '20 at 23:16

$$A\sin{\theta} + B\cos{\theta} = C$$ use the identity $$\cos\theta= \pm \sqrt{1-\sin^2\theta}$$ so $$A\sin{\theta} \pm B\sqrt{1-\sin^2\theta} = C$$ $$ \pm B\sqrt{1-\sin^2\theta} = C-A\sin{\theta}$$ $$ B^2(1-\sin^2\theta) = (C-A\sin{\theta})^2$$ let $x=\sin{\theta}$ $$ B^2(1-x^2) = (C-Ax)^2$$ $$B^2-B^2x^2=C^2-2ACx+A^2x^2$$ $$(A^2+B^2)x^2-(2AC)x+(C^2-B^2)=0$$ then use the quadratic formula to solve it and then complete the solution to find the $\theta$ values

Matthew H. · Answer 3 · 2020-09-29T00:15:36.190

0

If either $A$ or $B$ equals zero we can use basic trigonometry to solve the equation, so let's assume neither $A$ nor $B$ is zero. Under this assumption we can write $$A\cos(\theta)+B\sin(\theta)=\alpha\cos\Big(\theta -\arctan(B/A)\Big)$$ where $$\alpha=A\cos(\arctan(B/A))+B\sin(\arctan(B/A))$$ Can you continue from here?

edited Sep 29 '20 at 00:15

answered Sep 29 '20 at 00:06

Matthew H.

9,191

score 0 · Answer 4 · answered Sep 29 '20 at 12:29

Here's a general guide and explanation for problems of your type:

If we have an expression, $A\sin{x}+B\cos{x}$, let us assume it can be written in the form $R\sin(x+\alpha)$ Now to see if we can find values for $R$ and $\alpha$ in terms of $A$ and $B$. Using the compund angle formulae, also known as the addition formulae: $$R\sin(x+\alpha)=R\sin{x}\cos{\alpha}+R\sin\alpha\cos x=A\sin{x}+B\cos{x}$$ So we have $$R\cos\alpha=A,R\sin\alpha=B$$ So dividing the second equality by the first: $$\tan\alpha=\frac{B}{A}$$ meaning we can find $\alpha$ in terms of $A$ and $B$, as we wanted. Now, to find $R$: Squaring the $2$ equalities above we have $$R^2\cos^2\alpha+R^2\sin^2\alpha=R^2(\cos^2\alpha+\sin^2\alpha)=R^2=A^2+B^2\implies R=\sqrt{A^2+B^2}$$ So, to finish off by recapping what we have learnt: $$\tan\alpha=\frac{B}{A},~R=\sqrt{A^2+B^2}$$ Try applying that to your question. I hope that was helpful :)

How to solve $A\sin{\theta} + B\cos{\theta} = C$?

4 Answers4