Proving a "crucial trigonometry formula"

Question

So I was taught that to solve the equation: $$a\sin{x}+b\cos{x}=c$$ I would have to change it's form to $$k\cos{(x-\alpha)}=c$$ and solve it where $k^2=a^2+b^2$ and $\tan{\alpha}=\frac{a}{b}$

But throughout the practice questions, I've noticed that majority of them is solvable without the use of such formula. For example: $$\sin{x}+\sqrt{3}\cos{x}=1$$ $$2\left(\frac{1}{2}\sin{x}+\frac{1}{2}\sqrt{3}\cos{x}\right)=1$$ $$2\left[\sin{x}\sin\left(\frac{\pi}{6}\right)+\cos{x}\cos\left(\frac{\pi}{6}\right)\right]=1$$ $$2\left[\cos\left(x-\frac{\pi}{6}\right)\right]=1$$

But I can't find a way to prove it since it's considered the "wrong" method. Any sort of hints would help a ton. Thanks in advance!

Answer 1

You are showing twice the same method. Here is an alternative. WLOG, $a^2+b^2=1$.

$$a\sin x+b\cos x=c,$$

$$a\sin x=c-b\cos x,$$ $$a^2(1-\cos^2x)=c^2-2bc\cos(x)+b^2\cos^2(x)$$ and $$\cos^2x-2bc\cos(x)+c^2-a^2=0.$$

This is a quadratic equation in $\cos x$, giving

$$\cos x=bc\pm a\sqrt{1-c^2}.$$

From this,

$$\sin x=\frac{c-b\cos x}{a}=ac\mp b\sqrt{1-c^2}.$$

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This method is attractive when you only need the cosine and/or the sine, as it does not involve trigonometric functions.

Answer 2

Your method works when “notable” angles are involved. It's not difficult to find the general way. Since you want $$ a\sin x+b\cos x=k\cos(x-\alpha) $$ for every $x$, you can expand the right-hand side to $$ k\cos x\cos\alpha+k\sin x\sin\alpha $$ and you'll be satisfied if $$ k\sin\alpha=a,\qquad k\cos\alpha=b $$ This requires $$ k^2=a^2+b^2 $$ so you can choose $k=\sqrt{a^2+b^2}$ and $$ \cos\alpha=\frac{b}{\sqrt{a^2+b^2}}, \qquad \sin\alpha=\frac{a}{\sqrt{a^2+b^2}} $$ Note that this implies $\tan\alpha=a/b$, at least if $b\ne0$, but you cannot in general use $\alpha=\arctan(a/b)$, unless you're wanting to have $k<0$.