# 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.

• The use of displaystyle in the title is strongly depreciated. – Sangchul Lee Apr 2 '13 at 15:40
• Sorry I'm using a program that helps me to type these equations out – seeker Apr 2 '13 at 15:42