Find the values of $R$ and $\alpha$ in the identities below, given that $R>0$ and $\alpha$ is an acute angle.
$$\sqrt{3}\cos{\theta}-\sin{\theta}=R\cos(\theta+\alpha)$$
I'm a bit confused by this task.
How should I start? I have $$ \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b). $$
If I square anything, I can use the trig identity $$ \sin^2(x) + \cos^2(x) =1. $$
Do use the summation formula.
$$\sqrt3\cos\theta-\sin\theta=R\cos(\theta+\alpha)=R(\cos\theta\cos\alpha-\sin\theta\sin\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta.$$
Then you identify and solve $$R\cos\alpha=\sqrt3,\\R\sin\alpha=1.$$
Here's an alternative approach to those above which has a lot of geometric intuition:
Consider the vectors $u = (\cos\theta, \sin\theta)$, $v = (\sqrt{3}, -1)$. The vector $v$ makes an angle of with the positive $x$ axis of $\arctan(-1/\sqrt{3}) = -\pi/6$.
Then the original expression is the inner product between $u$ and $v$,
$$\sqrt{3}\cos\theta-\sin\theta = u \cdot v = \| u \| \ \| v \| \cos\alpha$$
where $\alpha$ is the angle between $u$ and $v$, namely $\alpha = \theta + \pi/6$. Also $\| u \| = 1$ and $\| v \| = \sqrt{3 + 1} = 2$.
Therefore
$$\sqrt{3}\cos\theta-\sin\theta = 2 \cos(\theta + \pi/6)$$
You're on the right track: $$R \cos(\theta+b)=R(\cos \theta\cos b -\sin \theta\sin b)=\sqrt3\cos\theta-\sin\theta$$ Thus $R\cos b=\sqrt3$ and $R\sin b=1$. Then $(R\cos b)^2+(R\sin b)^2=R^2=4$ so $R=2$ and $b=\arccos(\sqrt3/2)$.
