Welcome to compound angle formulae
In this stage we derive the compound angle formulae for \(\sin(A+B)\) and \(\cos(A+B)\) using rotations of vectors in the unit circle.
Take your time with each page. Work on the questions before moving to the answers!
Dr Brian Brooks
Mathematics InSight
Compound angle formulas: there are so many ways of proving these, but almost all these ways only work for a more or less narrow range of values. They also often shed little if any light on what is actually going on — they are technical proofs rather than intuitive proofs.
That's why I like the proof I'm about to show you so much: it is completely general, and it gives you a strong sense of what's really going on.
The idea is very simple: rotating a vector is the same as adding its rotated components.
Try my proof and see what you think for yourself!
Write this vector as a column vector
\[\binom{\cos\theta}{\sin \theta}\]
Now write this rotated vector as a column vector
\[\binom{\cos(\theta+\varphi)}{\sin (\theta+\varphi)}\]
Write the two yellow vectors as column vectors.
\[\binom{\cos\theta}{0}\] \[\binom{0}{\sin\theta}\]
After the rotation, what is the sum of the two yellow vectors?
\[\binom{\cos(\theta+\varphi)}{\sin (\theta+\varphi)}\]
You already know perfectly well that the unit vector \(\displaystyle\binom{1}{0}\) when rotated through the angle \(\varphi\) is \(\displaystyle\binom{\cos\varphi}{\sin\varphi}\).
What is the rotated yellow vector?
Here, the length of the vector is \(\cos\theta\), so the rotated vector is \[\cos\theta\binom{\cos\varphi}{\sin\varphi}\]
That's all very well, but what about cases where \(\cos\theta\) is negative?
Here is such a situation. Now the length of the yellow vector is positive, but \(\cos\theta\) is negative.
What is the rotated yellow vector this time?
Here, \(\cos\theta < 0\) and both components of the yellow vector are negative, so if the yellow vector is
\[\cos\theta\binom{\cos\varphi}{\sin \varphi}\]
then the signs of the vector work out as they should. That is, the yellow vector formula is still the same as when \(\cos\theta\) was positive. What a stroke of luck!
Check that the signs are right no matter the size of \(\varphi\).
Here, \(\theta > \pi\) and \(\cos\theta\) is still negative.
Once again, check that the signs are right no matter the size of \(\varphi\).
Rotating the vertical component is harder. We'll start with the unit vector in the vertical direction.
What is the rotated vertical unit vector?
The two components of the rotated vertical unit vector are the same as the two components of the the rotated horizontal unit vector only reversed and with one change of sign.
\[\binom{-\sin\varphi}{\phantom{-}\cos\varphi}\]Take the time to make sure you understand this.
Once again, check that the signs are right no matter the size of \(\varphi\).
Now we know what happens to the unit vertical vector when it rotates, but that's not what we are dealing with.
Instead, we are rotating the vertical component of a vector, whose magnitude is, in general, less than \(1\).
What is the rotated vertical component?
The magnitude of the vertical component is \(\sin\theta\), so the rotated vector is
\[\sin\theta\binom{-\sin\varphi}{\phantom{-}\cos\varphi}\]\(\sin\theta\) here is positive, so this vector is correct no matter the size of \(\varphi\).
That's all very well, but what about cases where \(\sin\theta\) is negative?
Here is such a situation. Now the length of the yellow vector is positive, but \(\sin\theta\) is negative.
What is the rotated yellow vector this time?
Here, \(\sin\theta < 0\) and the \(x\) component of the yellow vector is negative, so if the yellow vector is
\[\sin\theta\binom{-\sin\varphi}{\phantom{-}\cos \varphi}\]
then the signs of the vector work out as they should. That is, the yellow vector formula is still the same as when \(\sin\theta\) was positive. Another stroke of luck!
Check that the signs are right no matter the size of \(\varphi\).
Now we can write the blue vector in two ways.
Look what happens!
Well done!
You've now seen how the compound angle formulae arise naturally from rotating vectors in the unit circle.
Dr Brian Brooks
Mathematics InSight