The question asks me to solve $\displaystyle 7\cos x +6\sin x = 2$

So I put it into the form $\displaystyle r\sin (x + \alpha)$:

$\displaystyle 7\cos x +6\sin x \equiv r\sin (x + \alpha) \equiv r\sin x\cos\alpha + r\cos x\sin\alpha$

Then I equate the coefficients:

of cos x: $\displaystyle 7=r\sin\alpha $ [1]

of sinx: $\displaystyle 6=r\cos\alpha$ [2]

Divide [1] by [2] to get:

$\displaystyle \frac{7}{6}=tan\alpha \Rightarrow \alpha = 49.4^\circ$

Square and add [1] and [2] to get:

$\displaystyle 7^2+6^2=r^2(\sin^2\alpha+\cos^2\alpha) \Rightarrow r=\sqrt{85}$

$\displaystyle \therefore 7\cos x+6\sin x \equiv \sqrt{85}\sin(x+49.4^\circ)$

Except that last statement doesn't seem to be true, according to

the answers in my book and when I have a look at the two curves in a

graphing application. I tried $\displaystyle \sqrt{85}\sin(x-49.4^\circ)$ in the graphing app and

that seems to be correct...

but why??

What have I done wrong? Why am I getting a positive value for

alpha when I should be getting the same number but negative?

The only way I can see to get -49.4 would be by introducing a minus

sign when dividing [1] by [2]. But I don't see how to justify that.

Have I made a simple error? I can't spot it ...

Any help would be much appreciated, thanks in advance.