The addition formulae can be used to solve equations of the form:

(a \cos{θ} + b \sin{θ} = c )

Consider the expansion of (R\sin{(θ + α)} ):

(R\sin{(θ + α)} = R\sin{θ}\cos{α} + R\cos{θ}\sin{α} )

By matching terms in the expansion with the equation, we form a pair of simultaneous equations:

(R\sin{θ}\cos{α} = b \sin{θ} \implies R\cos{α} = b )
(R\cos{θ}\sin{α} = a \cos{θ} \implies R\sin{α} = a )

Solving for (R) involves squaring both equations and adding to eliminate (α):

(R^2\cos2{α} + R^2\sin^2{α} = b^2 + a^2)

(\implies R^{2\cancel{(\cos2{α}} + \sin^2{α})} = b^2 + a^2)

(\implies \boxed{R = \sqrt{b² + a^2}} )

Solving for (α) involves dividing the equations and eliminating (R):

(\dfrac{\cancel{R}\sin{α}}{\cancel{R}\cos{α}} = \dfrac{a}{b} )

(\implies \tan{α} = \dfrac{a}{b} )

(\implies \boxed{α = \arctan{\dfrac{a}{b}}} )

The original equation (a \cos{θ} + b \sin{θ} = c ) can therefore be rewritten as (R\sin{(θ + α)} = c ) and solved the usual way.

This is bnot/b given in the formula book. You are not expected to memorise the result, but to be able to derive it.

Tip: Variations of this method will work for other forms of the addition formulae, i.e. (R\sin{(θ-α)}), (R\cos{(θ+α)}) and (R\cos{(θ-α)}). Questions will usually tell you which version of the addition formula to use (but not always).

[b]ua cosθ + b sinθ[/u]/b

(a \cos{θ} + b \sin{θ} = R\sin{(θ + α)} )

where (R = \sqrt{b² + a^2}) and (\tan{α} = \dfrac{a}{b})