# Proving a "crucial trigonometry formula"

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!

• Your resolution method is exactly the one that leads to the general formula. For this reason, it is in no way wrong. Commented Aug 12 at 11:28
• You've become familiar enough with certain trig values to "notice" when certain coefficients are multiples of the sine & cosine of some angle, which causes you to "notice" that an expression is a multiple of an angle-addition (or -subtraction) identity. This is truly laudable ... & very convenient. The general formula is there for when you aren't (or can't be or won't bother to be) so clever. This is true of many formulas. For instance, lots of quadratic eqns are solvable by "notice"-able factoring; the Quadratic Formula provides a strategy that always works, without requiring cleverness.
– Blue
Commented Aug 12 at 12:26

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,

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

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

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 $$k<0$$.