How to find α in this Reduction Identity

Question

In the reduction identity :

m sin θ + n cos θ = √(m² + n²) sin(θ + α)

I am having trouble with determining the value of α. Here is an example.

Problem : -7 sin θ - 24 cos θ
m = -7
n = -24
Using the above formula :
√(-7² + -24²) sin(θ + α) = 25 sin(θ + α)

From here, I use the following identities to attempt to determine α

sin α = n / √(m² + n²)

cos α = m / √(m² + n²)

sin α = -24/25, α = -74°
cos α = -7/25, α = 106°

At this point, I have two possible values for α. My textbook states, "α is the smallest possible positive value that satisfies both of these conditions," and lists the value of α as 254°. I'm a bit confused. How did they arrive to that conclusion, and what steps can I take to solve the problem?

Where did a and b come from, are those supposed to be m and n? — Collapse, Feb 03 '17 at 17:36
Hint: see this page .http://math.stackexchange.com/questions/1384479/solve-the-trig-equation-cos-theta-%E2%88%92-sin-theta-1/1384530#1384530 — Khosrotash, Feb 03 '17 at 17:37
$${\color{Red}{a sinx+ b cos x= \frac{|a|}{a} \sqrt{a^2+b^2}sin (x+\alpha)\tan \alpha=\frac{b}{a}} }$$ — Khosrotash, Feb 03 '17 at 17:38

grand_chat · Accepted Answer · 2017-02-03T18:06:08.830

The equation $\sin\alpha=-24/25$ is satisfied not only by $\alpha=-74^\circ$ (plus multiples of $360^\circ$); there's another family of solutions, namely $\alpha=180^\circ+74^\circ$ (plus multiples of $360^\circ$). (Think about where on the unit circle you can get a sine of $-24/25$; there will be two places, equally distant from $-90^\circ$.)

Similarly there's a second family of solutions to the equation $\cos\alpha=-7/25$, namely $\alpha=-106^\circ$ (plus multiples of $360^\circ$.) (The two places where you can get a cosine of $-7/25$ will be equally distant from $180^\circ$.)

Putting these together you'll find $\alpha=254^\circ$ is the smallest positive $\alpha$ that meets the requirements.

How to find α in this Reduction Identity

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