How to solve $a \cos \alpha + b \sin \alpha = c$ for $\alpha$?

Question

I'm solving a physics problem and I came down to solving an equation of the form $$a \cos \alpha + b \sin \alpha = c$$

Can someone help me to solve this? Thanks in advance!

Answer 1

Let $r=\sqrt{a^2+b^2}$, then there is an angle $\theta$ with $a=r\cos\theta, b=r\sin\theta$. Now the equation becomes $$r\cos\theta\cos\alpha+r\sin\theta\sin\alpha=c\\ \cos(\alpha-\theta)=c/r$$

Answer 2

Multiply both sides by $\frac{1}{\sqrt{a^2+b^2}}$ And then use sine formula for $\sin(a+b)$

Answer 3

HINT: $$\frac{a}{\sqrt{a^2+b^2}}\cos\alpha+\frac{b}{\sqrt{a^2+b^2}}\sin\alpha=\frac{c}{\sqrt{a^2+b^2}}$$ Now there's $\beta$, such that : $$\cos\beta\cos\alpha+\sin\beta\sin\alpha=\frac{c}{\sqrt{a^2+b^2}}$$ or $$\cos(\alpha-\beta)=\frac{c}{\sqrt{a^2+b^2}}$$

Answer 4

HINT: Express $$\sin\alpha=\frac{2t}{1+t^2},\cos\alpha=\frac{1-t^2}{1+t^2}$$ $$t=\tan(\alpha/2)$$