There’s a very typical question in C3 papers that looks something like:

“Express $3 \sin (x) + 5 \cos (x)$ in the form $R \sin (x + \alpha)$.”

(Sometimes it’s a different trig function, or it may have a minus sign in it, but the same principle works for any type of question like this, which I’m going to call an R-sin-alpha question).

There are students about who would panic at this, but you don’t need to be one of them. Everything you need to do this is in your formula book, and it’s a completely mechanical process once you know what you’re doing.

The trick is to start with the $R \sin (x + \alpha)$ thing. You’re given a formula for $\sin(A+B)$, which is $\sin(A) \cos (B) + \sin (B) \cos (A)$. ((Obviously, if you’re doing a different trig function, use the appropriate formula. It’s not just R-sin-alpha, it can be R-cos-alpha, too.))

That means $R \sin (x + \alpha) = R ( \sin (x) \cos (\alpha) + \sin (\alpha) \cos (x) ) \\= R \sin (x) \cos (\alpha) + R \sin (\alpha) \cos (x)$.

Now comes the clever bit. That expression has to be EXACTLY the same as $3 \sin (x) + 5 \cos (x)$ - you need the same number of $\sin(x)$s and the same number of $\cos(x)$s. There are $3 \sin(x)$s in the first expression, and $R \cos(\alpha)$ $\sin(x)$s in the second, so $3 = R \cos (\alpha)$. In the same way, $5 = R \sin (\alpha)$.

Two equations? Two unknowns? That looks like a simultaneous equation to me. It’s a bit tricky because they’re non-linear, but don’t let a thing like that bother you: divide the $R \sin(\alpha)$ one by the $R\cos(\alpha)$ one and the $R$s cancel to leave you with $\tan(\alpha) = \frac{5}{3}$. You can throw that in your calculator and get an answer out - probably somewhere about 1 radian, let me check. (1.03, it says here.)

What’s that? No, degrees are rubbish. Don’t use them unless you’re explicitly told to.

Once you have the angle, you have a couple of options: you can choose to do it the ‘proper’ way, which involves squaring the two equations and adding them up (you end up with $R^2 \cos^2(\alpha) + R^2 \sin^2(\alpha) = 3^2 + 5^2 = 34$; you can use a famous trig identity to get rid of the $\cos$ and the $\sin$ to leave you with $R = \sqrt{34}$.

Alternatively, since you know what $\alpha$ is, you can just work from one of the equations - $3 = R \sin (\alpha)$, so $R = \frac{3}{\sin (\alpha)}$. Do $\frac{3}{\sin(Ans)}$ on your calculator - don’t type it all in again, that’s a) a waste of effort and b) inaccurate - and you’ll get about 5.83. It’s worth checking if that’s an exact square root, because square roots are nearly always preferable to decimals - oh look! it’s $\sqrt{34}$.

So, you can finish the R-sin-alpha question by writing $3 \sin (x) + 5 \sin (x) = \sqrt{34} \sin (x + 1.03)$. There’s three or four marks for you right there. You’re welcome.

* Updated 2016-12-30 to square the trig functions. Thanks to James for spotting the error!