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In the proof of Transforming $a\cos\left(\, x\,\right)+b\sin\left(\, x\right)$ to $r\cos\left(\,\phi - x\,\right)$

\begin{align} a\cos\left(\, x\,\right) + b\sin\left(\, x\,\right) &=\,\sqrt{\,a^{2} + b^{2}\,}\, \left[\,\frac{a}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\cos\left(\, x\,\right) +\frac{b}{\,\sqrt{\, a^{2} + b^{2}\,}\,}\,\sin\left(\, x\,\right)\,\right]\, \\[2mm]&=\,\sqrt{\,a^{2} + b^{2}\,}\,\left[\, \cos\left(\,\phi\,\right)\cos\left(\, x\,\right) + \sin\left(\,\phi\,\right)\sin\left(\,x\,\right) \,\right] \\[2mm]&=\,\sqrt{\, a^{2} + b^{2}\,}\,\cos\left(\,\phi - x\,\right) \end{align}

why did we factor out $\,\sqrt{\, a^{2} + b^{2}\,}\,$ where did this idea came from ?. It seems out of the blue.

Felix Marin
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cosacosb
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    In trigonometry a lot of things revolve around right triangles with hypotenuse equal to $1$. A triangle with legs $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ will have hypotenuse $1$ by the Pythagorean theorem. – Arthur Dec 21 '14 at 11:36
  • My answer on this question could be of interest to you; it doesn't explain this point in particular, but I think it's clear enough so that it would resolve your question. – Milo Brandt Dec 22 '14 at 23:56

2 Answers2

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Start at "the other end" of the problem. If we want $$a\cos x+b\sin x=r\cos(\phi-x)$$ and we expand the right hand side, then what we are looking for is $$a\cos x+b\sin x=r\cos\phi\cos x+r\sin\phi\sin x\ .$$ If this is to be true for all values of $x$ then we need $$a=r\cos\phi\quad\hbox{and}\quad b=r\sin\phi\ .$$ Squaring and adding these equations, $$a^2+b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2$$ and so $$r=\sqrt{a^2+b^2}\ .$$ That's where the idea comes from to factor out $\sqrt{a^2+b^2}$.

David
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The idea behind the proof comes from noting that

$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\dfrac{b}{\sqrt{a^2+b^2}}\right)^2=1.\tag1$$

This is analogous to the Pythagorean identity

$$\cos^2\phi+\sin^2\phi=1.\tag2$$

So for all $a$ and $b$ we can get an expression that satisfies $(2)$ by just factoring out $\sqrt{a^2+b^2}$, but in what way will this be useful? From $(1)$ and $(2)$ we can say that there exists an angle $\phi$ such that

$$\cos\phi=\dfrac{a}{\sqrt{a^2+b^2}}\qquad\sin\phi=\dfrac{b}{\sqrt{a^2+b^2}},$$ or $$\cos\phi=\dfrac{b}{\sqrt{a^2+b^2}}\qquad\sin\phi=\dfrac{a}{\sqrt{a^2+b^2}}.$$

All of this will permit us to transform $a\cos x+b\sin x$ into an expression of the form

$$\text{constant}\cdot(\sin\phi\cos x\pm\cos\phi\sin x),$$ or $$\text{constant}\cdot(\cos\phi\cos x\pm\sin\phi\sin x).$$

Those two are the result of the angle-addition formulas, which will enable us to write it in the form $r\cos(\phi\pm x)$ or $r\sin(\phi'\pm x)$.

In conclusion, the proof boils down to determining how to transform the coefficients of $\cos$ and $\sin$ such that they are exactly equal to the $\cos$ and $\sin$ of an angle. The rest follows from the angle-addition formulas.

Workaholic
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