How do you simplify something of the form Acos(x) + Bsin(x) ?
This sort of question is quite common, and a *little* bit complicated, so the explanation is quite long! But it's a great one to understand. There are a few tricks you need to know to be able to do it. The first thing you need to know is what you can simplify it to.
There are two similar formats you might be asked to simplify it into; these are Csin(x+e), or Ccos(x+f). The fact that you can simplify this sum into a single sinusoidal (sinusoidal = sine or cosine) function means that the new wave has the same shape as either of the original waves, but it will have been stretched and laterally translated (shifted sideways). The C represents the amplitude of the new wave (how high and low it stretches), and e or f represents the lateral translation (sideways offset from a normal sine or cosine wave).
To explain the method for working out C and e or f, it is probably easiest to use an example! Let's start with:
3cos(x) - sin(x)
and try and get to the new form Csin(x+e). This means A = 3, and B = -1
Let's start by equating our original expression with the target expression: 3cos(x) - sin(x) = Csin(x+e)
The first trick is to expand the right hand side using the general rule sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)
Remembering to multiply everything by C, we get: 3cos(x) - sin(x) = Csin(x)*cos(e) + Ccos(x)*sin(e)
Before we do the second trick, I'm going to rearrange the equation above slightly:
[3]*cos(x) + [-1]*sin(x) = [Csin(e)]*cos(x) + [Ccos(e)]*sin(x)
The second, and strangest, trick involves realising that both sides of the equation are in the same format, that is [something]*cos(x) + [something]*sin(x) . We can actually equate both pairs of somethings because of this! This gives us two new equations:
3 = Csin(e) , and -1 = Ccos(e)
What this trick has done is gotten rid of x from the equations and let only C and e, the two variables we want to find. Now we can use these two new equations like simultaneous equations to work out C and e. Wonderful!
The easiest way to work out C and e now is to find a way to eliminate one of them from the equations. It doesn't matter which one you try to eliminate first, but normally I choose to eliminate e first.
To do so, first square both of the simultaneous equations to give 9 = C2sin2(e) and 1 = C2cos2(e) . Then you can make use of a special trigonometric identity, which is sin2(x) + cos2(x) ≡ 1, by adding both equations together to give;
10 = C2[sin2(e) + cos2(e)], which gives 10 = C2. This means C = sqrt(10) . Almost there!
Now we have to work out e. There are a few ways to do this, and unfortunately, all of them involve some pitfalls! I'll show you what I think is the easiest way to go about this.
Returning to the two simultaneous equations, this time, divide the first (the one with the sin term) by the second (the one with the cos term), to give 3 / -1 = Csin(e) / Ccos(e), which gives -3 = tan(e)
Now we can use the inverse tan function (tan-1 or arctan on your calculator) to solve for e; e = tan-1(-3)
Now this is where the possible pitfall appears. Your calculator will probably give the answer e = -1.25 (using the radians setting instead of degrees). However, one of the special things about inverse trig functions is that they actually have multiple solutions, and your calculator can only show one! The reason I chose this method is because working out the other possible solutions is quite easy for the inverse tan function, simply add or subtract multiples of pi (π) from -1.25.
So, how do you know what the right solution is? There are always two possible solutions; your calculator output (-1.25) or that output plus pi (-1.25+π). First let's test -1.25. This would make our very final solution;
3cos(x) - sin(x) = sqrt(10)*sin(x-1.25)
Is this correct? Let's see what happens at x=0, when cos(x)=1 and sin(x)=0;
3 - 0 = sqrt(10)*sin(-1.25)
If you trying inputting sqrt(10)*sin(-1.25) into your calculator (in radians), you should get about -3, not +3, which means this solution is incorrect. So now we should go back and try e=-1.25+π, which makes our final solution;
3cos(x) - sin(x) = sqrt(10)*sin(x-1.25+π)
If you apply the same test at x=0, you will find the right hand side does give +3, which shows this solution is correct. Which means...
HOORAY! We're done!
----------------some extra notes----------------
If you want to convince yourself that it is possible to express the sum of a sine and cosine term as a single function, try plotting it on Wolfram Alpha (e.g. " plot y=3cos(x)-sin(x) ", you'll see it has the same shape!
However, this only works when you have the same thing inside the sine and cosine brackets. For instance, if you had 3cos(x) - sin(2x), it wouldn't be possible.
If you're asked for the simplified form Ccos(x+f), you'll need to know the expansion rule for cos, which is;
cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)
from which the same principle can be applied.
A **very helpful shortcut** is to notice that the resultant amplitude (C) can always be easily worked out using the rule C2 = A2 + B2 ! It's good to use this as a double check once you've worked it out normally.
***It's also worth saying that, in an exam situation, even if you don't remember the little things like maybe needing to add pi to the bracket, you could still easily get most or even all of the marks! So don't panic if this seems very complicated, I'm just trying to make my explanation as thorough as possible!***
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