<s>[INST] Find the value of $\alpha $ given $2\sin\theta -\sqrt 5 \cos \theta \equiv - 3\cos (\theta + \alpha )$ Given:
$$2\sin\theta -\sqrt 5 \cos \theta  \equiv  - 3\cos (\theta  + \alpha ),$$
where $$0 <\alpha  < 90^\circ, $$
find $α.$
The issue I have with this question is the $-3$ on the right hand side, it really complicates things and I don't know how to deal with it.
When I usually come across equations of the form $$a\sin \theta  + b\cos \theta $$ it's a relatively straight forward process of converting them into another equation of the form
$$\sqrt {{a^2} + {b^2}} \left({a \over {\sqrt {{a^2} + {b^2}} }}\sin \theta  - {b \over {\sqrt {{a^2} + {b^2}} }}\cos \theta \right)$$
where $$\cos \alpha  = {a \over {\sqrt {{a^2} + {b^2}} }}$$
and $$\sin \alpha  = {b \over {\sqrt {{a^2} + {b^2}} }}$$
(or some other variant of this).
I cant wrap my head around this, specifically. If
$$2\sin\theta -\sqrt 5 \cos \theta  \equiv  - 3\cos (\theta  + \alpha ),$$
this must mean the square root expression prefixing the cosine addition identity must be $-3$, as
$$\sqrt {{a^2} + {b^2}} \cos(\theta  + \alpha ) \equiv  - 3\cos(\theta  + \alpha ).$$
I am aware the square root operation outputs negative values I just dont know how to deal with that in this instance.
I hope this doesn't read too convoluted, thank you.
 [/INST] You can use the so-called double angle formula. Notice that
$$-3\cos(\theta+\alpha) \equiv -3\cos\theta\cos\alpha + 3 \sin\theta\sin\alpha $$
Now you can compare coefficients. You want $-3\cos(\theta+\alpha) \equiv 2\sin\theta - \sqrt{5}\cos\theta$, and so you need to find an $\alpha$ for which $3\sin\alpha = 2$ and $3\cos\alpha = \sqrt{5}$. It follows that
$$\frac{3\sin\alpha}{3\cos\alpha} \equiv \tan\alpha = \frac{2}{\sqrt{5}}$$
You $\alpha$ is then $\arctan(2/\sqrt{5}) \approx 41.8^{\circ}$.
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