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Could someone help me with the steps for solving the below equation $$a \sin\theta + b \cos\theta = c$$ I know that the solution is $$\theta = \tan^{-1} \frac{c}{^+_-\sqrt{a^2 + b^2 - c^2}} - \tan^{-1} \frac{a}{b} $$ I just can't figure out the right steps to arrive at this solution.

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4 Answers 4

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here is a trick: write $$\frac{a}{\sqrt{a^2+b^2}}\sin(\theta)+\frac{b}{\sqrt{a^2+b^2}}\cos(\theta)=\frac{c}{\sqrt{a^2+b^2}}$$ Setting $$\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$$ and $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$ then you will get $$\sin(\phi+\theta)=\frac{c}{\sqrt{a^2+b^2}}$$ so $$\theta=\arcsin\left(\frac{c}{\sqrt{a^2+b^2}}\right)-\phi$$

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  • $\begingroup$ I'd love to figure this out, but I can't seem to get from here to the solution above... $\endgroup$ Commented Feb 14, 2018 at 17:21
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Well, we have:

$$\text{a}\cdot\sin\left(x\right)+\text{b}\cdot\cos\left(x\right)=\text{c}\tag1$$

Substitute $\text{y}=\tan\left(\frac{x}{2}\right)$, so $\sin\left(x\right)=\frac{2\cdot\text{y}}{1+\text{y}^2}$ and $\cos\left(x\right)=\frac{1-\text{y}^2}{1+\text{y}^2}$:

$$\text{y}^2-\frac{\text{b}-\text{c}}{\text{b}+c}-\frac{2\cdot\text{a}\cdot\text{y}}{\text{b}+\text{c}}=0\tag2$$

Solving for $\text{y}$, gives:

$$\text{y}=\pm\sqrt{\frac{\text{a}^2}{\text{b}+\text{c}}+\frac{\text{b}-\text{c}}{\text{b}+\text{c}}}+\frac{\text{a}}{\text{b}+\text{c}}\tag3$$

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You can rewrite the expression on the left side: $$ a\cdot\sin(\theta)+b\cdot\cos(\theta)= A\sin(\theta+\tau)= A\cdot\sin(\theta)\cos(\tau)+A\cdot\cos(\theta)\sin(\tau) $$

$$A\cdot\cos(\tau)=a$$ $$A\cdot\sin(\tau)=b$$

The paramteres of the rewritten form: $$A=\sqrt{a^2+b^2}$$ $$\tan(\tau)=\frac{b}{a}$$

You can express $\tan(\theta+\tau)$ using $\cos(\theta+\tau)=^+_-\sqrt{1-\sin^2(\theta+\tau)}$.

You will get the result in the form you want it.

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Recall the sum formula for sine $$A\sin(\theta+\alpha)=A\sin(\theta)\cos(\alpha)+A\cos(\theta)\sin(\alpha)$$

Equating this with the L.H.S gives

$$a\sin(\theta)+b\cos(\theta)=A\cos(\alpha)\sin(\theta)+A\sin(\alpha)\cos(\theta)$$

so we get the system

$$a=A\cos(\alpha)\\b=A\sin(\alpha)$$

and so $A=\sqrt{a^2+b^2}$.

Recall that we're defining some angle $\alpha$ such that both $\cos(\alpha)=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$, so $\alpha$ is unique over one rotation and falls within a quadrant that depends on the signs of $a$ and $b$. You should be able to convince yourself that $\alpha$ takes the value

$$\alpha=\begin{cases} \arctan(b/a), &a\gt0\\ \arctan(b/a)+\pi, &a\lt0 \end{cases}$$

So for $a>0$, we get

$$ a\sin(\theta)+b\cos(\theta)=\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a)) $$

and from the identity $\sin(\theta+\pi)=-\sin(\theta)$, then for $a<0$ we get

$$ \sqrt{a^2+b^2}\sin(\theta+\arctan(b/a)+\pi)=-\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a)) $$

so we can conclude that in general

$$ a\sin(\theta)+b\cos(\theta)=\frac{a}{|a|}\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))\quad (a\neq 0) $$

For $-\sqrt{a^2+b^2}\leq c\leq\sqrt{a^2+b^2}$, we get

$$ \frac{a}{|a|}\sqrt{a^2+b^2}\sin(\theta+\arctan(b/a))=c\\ \implies\sin(\theta+\arctan(b/a))=\frac{|a|c}{a\sqrt{a^2+b^2}}\\ \implies\theta_1=\arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)-\arctan\left(\frac{b}{a}\right),\\ \theta_2=\pi-\arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)-\arctan\left(\frac{b} {a}\right) $$

Use the identity $\arcsin(x)=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$ with $x=\frac{|a|c}{a\sqrt{a^2+b^2}}$ to get

$$ \arcsin\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\right)=\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2}}\cdot\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2-c^2}}\right)\\ =\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right) $$

and conclude that the general solution becomes

$$ \theta_1=\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right)-\arctan\left(\frac{b}{a}\right)+2\pi k_1\\ \theta_2=-\arctan\left(\frac{|a|c}{a\sqrt{a^2+b^2-c^2}}\right)-\arctan\left(\frac{b}{a}\right)+\pi(2k_2+1) $$

where $k_1,k_2\in\mathbb{Z}$ and $a\neq 0$.

As desired.

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