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I'm asked to write $$f(x) = a\cdot\cos{x} + b\cdot\sin{x}$$ on the form $$A\cdot\sin{(x+d)}$$ So we need to find an A and a d to satisfy: $$A\cdot\sin{(x+d)} = a\cdot\cos{x} + b\cdot\sin{x}$$ $$A\cdot\sin{x}\cos{d} + A\cdot\sin{d}\cos{x} = a\cdot\cos{x} + b\cdot\sin{x}$$ we get $$A\cos{d} = b$$ $$A\sin{d} = a$$

This is a system of equations with 3 unknowns. Now, I recognize that each equation can be perceived as coordinates in the formal system of trigonometry. But what additional information does it actually provide us to make this traditionally unsolvable problem solvable? I have a hard time grasping this. How would you explain why it works to someone who knows nothing about trigonometry?

user361931
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2 Answers2

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Let me propose a somewhat geometric explanation. I hope you are familiar a bit with vectors.

Assume you have the standard orthonormal coordinate system $O\overrightarrow{e}_1\overrightarrow{e}_2$ (usually called the $x,y$ coordinate system, but here, we cannot call it like that, since in our case the label $x$ is used denote an angle and not a Cartesian coordinate). All vectors we work with are arrows attached to the origin $O$ of the coordinate system. So we are given a function $f(x) = a\cos(x) + b\sin(x)$. Then, define the two vectors $\overrightarrow{c} = a \, \overrightarrow{e}_1 + b \, \overrightarrow{e}_2$ and $\overrightarrow{v}(x) = \cos{(x)} \, \overrightarrow{e}_1 + \sin{(x)} \, \overrightarrow{e}_2$. These two vectors are like the two hands of a clock. The vector $\overrightarrow{v}(x)$ is a unit vector having angle $x$ with the horizontal coordinate axis determined by $\overrightarrow{e}_1$, and $\overrightarrow{v}(x)$ has length $|\overrightarrow{v}(x)| = \sqrt{\cos^2{x} + \sin^2{x}} = 1$. The other vector has length $|\overrightarrow{c}| = \sqrt{a^2+b^2}$. If you take the dot product between the two vectors you get $(\overrightarrow{c} \cdot \overrightarrow{v}(x)) = a\cos(x) + b\sin(x) = f(x)$ -- exactly the function you are given. Geometrically, this last dot product is the length of the orthogonal projection (the shadow) of vector $\overrightarrow{c}$ onto the line determined by the vector $\overrightarrow{v}(x)$. If $\theta$ is the angle between the two vectors $\overrightarrow{v}(x)$ and $\overrightarrow{c}$ and $A=|\overrightarrow{c}| = \sqrt{a^2+b^2}$ is the length of vector $\overrightarrow{c}$, then $$f(x) = a\cos(x) + b\sin(x) = (\overrightarrow{c}\cdot\overrightarrow{v}(x)) = |\overrightarrow{c}| |\overrightarrow{v}(x)| \cos{\theta} = |\overrightarrow{c}| \cos{\theta} = \sqrt{a^2+b^2} \cos{\theta} = A \cos{\theta}.$$ To calculate the angle $\theta$ between $\overrightarrow{v}(x)$ and $\overrightarrow{c}$, we can simply subtract from the angle $x$ between $\overrightarrow{v}(x)$ and the coordinate axis vector $\overrightarrow{e}_1$ the angle $\tilde{d}$ between $\overrightarrow{c}$ and the coordinate axis vector $\overrightarrow{e}_1$, that is $\theta = x - \tilde{d}$. Thus $$f(x) = a\cos(x) + b\sin(x) = (\overrightarrow{c}\cdot\overrightarrow{v}(x)) = A \cos{\theta} = A \cos{(x - \tilde{d})}.$$ Almost there. We just rewrite the angle $\tilde{d} =\pi/2 - d$ and obtain $$f(x) = a\cos(x) + b\sin(x) = A \cos{(x + d - \pi/2)} = A \sin{(x+d)}.$$

Futurologist
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Let's see how $C\sin(x + d)$ is equivalent to $A\cos (x) + B\sin(x)$. Define $C = \sqrt{A^2 + B^2}$. Then, $A\cos (x) + B\sin(x) = C\left[\frac{A}{C}\cos x + \frac{B}{C} \sin x\right]$. Since $|A/C|, |B/C| \leq 1$, $A/C = \sin d$ and $B/C = \cos d$ for some $d \in [0,2\pi)$. We have $C\left[\frac{A}{C}\cos x + \frac{B}{C} \sin x\right] = C\left[\sin d\cos x + \cos d \sin x\right] = C\sin(x + d)$, where all we have done in the last equality is used the identity $\sin\alpha\cos\beta + \cos\alpha\sin\beta = \sin(\alpha + \beta)$.

In other words, $C = \sqrt{A^2 + B^2}$ and $d = \arctan{A/B}$.

Alex Ortiz
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