A few questions regarding the cosine difference identity

Question

I've a few questions that stem from the proof given in my textbook regarding the cosine difference identity.

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The proof goes like this:

Let $\alpha$ and $\beta$ be angles plotted in standard position on the Unit Circle, where $\alpha \geqslant\beta $. (Ignore the $\alpha _0$ and $\beta _0$ in the diagram, just take them to mean $\alpha$ and $\beta$). Let $P$ be the point on the terminal side of $\alpha$ that lies on the Unit Circle, and $Q$ be the point the terminal side of $\beta$ that lies on the Unit Circle.

This gives:

Plotting the angle $\alpha -\beta$ in standard position gives:

Let $A$ be the point on the terminal side of $\alpha -\beta$ that lies on the Unit Circle, and $B$ be the point where the initial side of the angle $\alpha -\beta$ intersects the Unit Circle.

As the angles $POQ$ and $AOB$ are congruent this must mean the chords $PQ$ and $AB$ are also equal, and it is possible to set up an equivalance relationship between the two. The distance formula yields:

$$\sqrt {(\cos \alpha - \cos \beta )^2 + (\sin \alpha - \sin \beta )^2} = \sqrt {(\cos (\alpha - \beta ) - 1)^2 + (\sin (\alpha - \beta ) - 0)^2} $$

Squaring both sides to remove the square root:

$$(\cos \alpha - \cos \beta )^2 + (\sin \alpha - \sin \beta )^2 = (\cos (\alpha - \beta ) - 1)^2 + (\sin (\alpha - \beta ) - 0)^2$$

$$\cos ^2\alpha - 2\cos \alpha \cos \beta + \cos ^2\beta + \sin ^2\alpha - 2\sin \alpha \sin \beta + \sin ^2\beta = \cos ^2(\alpha - \beta ) - 2\cos (\alpha - \beta ) + 1 + \sin ^2(\alpha - \beta )$$

$$(\cos ^2\alpha + {\sin ^2}\alpha ) + (\cos ^2\beta + \sin ^2\beta ) - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = (\cos ^2(\alpha - \beta ) + \sin ^2(\alpha - \beta )) + 1 - 2\cos (\alpha - \beta )$$

$$2 - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = 2 - 2\cos (\alpha - \beta )$$

$$\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $$

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So that's how it was derived. My questions are:

1. As the point P (the point where the terminal side of the angle $\alpha$ intersects the Unit Circle) is in the second quadrant, where $x<0$ and therefore $\cos (\alpha)<0$ shouldn't the coordinates for the point $P$ be $(-\cos(\alpha), \sin(\alpha))$? Furthermore what would happen if the angle $\alpha$was a quadrant III angle where both $\sin (\alpha)<0 $ and $\cos (\alpha )<)$. Surely this would affect computations of the cosine difference identity some what? If not, then why? Why does it seem like its computed only for quadrant I angles? Why does that identity stand for all quadrants is what I'm asking essentially?

2. The cosine difference formula is derived by equating the distance\length of the two chords $PQ$ and $AB$, however we "square out" the square root inherent in the distance formula. What if I were to do: $\sqrt {2 - 2\cos (\alpha - \beta )} $ (which is the simplified version of the RHS of the equation before we manipulate it to give us the identity) Would this give me the distance between the two chords?

3. What difference would it make if angle $\alpha $ was smaller than $\beta $?

4. What would happen if the angle $(\alpha -\beta)$ was greater $180^\circ$ or $\pi$ radians?

EDIT: Clarification of question 1 & 2., added a question 3 and 4.

Answer 1

I will be posting some thoughts I have on the questions you have posted above. Feel free to ask any questions:

1. As the point P (the point where the terminal side of the angle $\alpha$ intersects the Unit Circle) is in the second quadrant, where $x<0$ and therefore $\cos (\alpha)<0$ shouldn't the coordinates for the point $P$ be $(-\cos(\alpha), \sin(\alpha))$? Furthermore what would happen if the angle $\alpha$was a quadrant III angle where both $\sin (\alpha)<0 $ and $\cos (\alpha )<)$. Surely this would affect computations of the cosine difference identity some what? If not, then why? Why does it seem like its computed only for quadrant I angles? Why does that identity stand for all quadrants is what I'm asking essentially?

Ok, that's a mouthful. :) I think @angryavian has answered the first part of your first question: i.e. $\cos \alpha$ may not be positive, hence denoting the $x$-coordinates of point $P$ as $\cos \alpha$ is not wrong. For example, in this case (shown in diagram), point $P$ is found in the second Quadrant. Then $\cos \alpha <0$ (just like you said), thus, we don't have to put a negative sign in front of $\cos \alpha $, otherwise, in this case, $- \cos \alpha >0$, which is incorrect.

Next, note that the proof does not restrict $\cos \alpha$ or $\sin \alpha$ to be in any particular quadrant, the illustration is simply one example. The condition mentioned was for $\alpha > \beta$ - this is intended to be WLOG, or without loss of generality as I will mentioned in response to Question 3 later on. Notice that the distance formula (using Pythagoras) calculates the length of chords $PQ$ and $AB$ by making use of "rise" (differences in $y$-coordinate) and "run" (differences in $x$-coordinate). Thus, the sign of $\sin \alpha$ or $\cos \alpha$ (as you mentioned "$\sin (\alpha)<0 $ and $\cos (\alpha )<)$") doesn't matter. I will show you an example to see what I meant:

Suppose in Quadrant III, $\sin (\alpha)<0 $ and $\cos (\alpha )<0)$, and let us suppose $\beta$ remains in Quadrant I (it doesn't matter), then $\sin (\beta), \cos(\beta)>0 $. Calculating the "rise", we use $\sin (\alpha)-\sin (\beta)$. This will result in a negative "rise". However, notice that in the distance formula, we square this expression, hence convert the negative "rise" into positive. This is in essence taking the absolute value of the "rise", hence giving the magnitude/side of the triangle (if you see the whole "rise" and "run" thing as finding the gradient using a gradient triangle method.) - since the side of a true triangle cannot be negative (negativity applies when we are considering vectors). The same applies for "run". Thus, back to your question, the sign of the $\cos, \sin$ doesn't matter because ultimately, we want to find the absolute value of the distance of the side of the triangle (either rise or run).

So, why does the formula hold for all quadrant? As I mentioned earlier, the signs of the $\cos, \sin$ doesn't matter when it comes to applying the distance formula for chord $PQ$ or $AB$.

2.The cosine difference formula is derived by equating the distance\length of the two chords $PQ$ and $AB$, however we "square out" the square root inherent in the distance formula. What if I were to do: $\sqrt {2 - 2\cos (\alpha - \beta )} $ (which is the simplified version of the RHS of the equation before we manipulate it to give us the identity) Would this give me the distance between the two chords?

Yes, I believe so.

3.What difference would it make if angle $\alpha $ was smaller than $\beta $?

Note that $\cos x = \cos(-x)$. This can be observed from the $\cos$ graph. Thus, if angle $\angle \alpha $ was smaller than $\angle \beta $, observe that $\cos (\alpha - \beta) = \cos(-(\alpha - \beta) ) = \cos(\beta - \alpha)$. Hence, formula still holds. Alternatively, you can also substitute $\beta for \alpha$ in the formula (I know this sounds weird) and $\alpha$ for $\beta$ if $ \angle \alpha < \angle \beta $.

4.What would happen if the angle $(\alpha -\beta)$ was greater $180^\circ$ or $\pi$ radians?

Refer to the Unit Circle ASTC (All positive, sine, tangent, cosine) as follows:

Consider $360° >$ α - β $> 180°$, then, when β is in the Quadrant I, α can be either in Quadrant III or IV; β is in the Quadrant II, α can be either in Quadrant IV or I (but if I we go back to the case when α < β. For both cases, notice if we minus $180$° from the angle α (α still $>$ than β), regardless of whether α was in Quadrant III or IV, its $\cos$ and $\sin$ both switched signs (from negative to positive and positive to negative).

This is aligned to the fact that if $x - y < 180°$, then, $\cos(x - y) = - \cos(x - y +180°)$. (not the same α and β are mentioned above). Using bijection, we can ensure that for every α - β > 180°, there is $x - y < 180$°, where x - y + 180° = α - β. E.g. We are interested in $α - β = 270$°, then there will be at least a pair of $x - y = 270 - 180 = 90$°, then $\cos(α - β) = - \cos(x - y +180°)$. This holds for all α, β.

Note: When $\alpha - \beta > 360°$ then since $cos(\alpha - \beta) = cos(\alpha - \beta + 360°)$ and that $\cos x = \cos (360°+x)$ and $\sin x = \sin (360°+x)$ for all x. Our formula holds for all angles.

Answer 2

1. If $\cos\alpha<0$ and $\sin\alpha>0$, then $(\cos\alpha,\sin\alpha)$ is in the second quadrant, which is what you want, right? $(-\cos\alpha,\sin\alpha)$ would be in the first quadrant.

2. No, $\sqrt{\cos(\alpha-\beta)}$ is not the length of the chord because you have manipulated the equations... consider this argument: \begin{align*} \sqrt{x}&=\sqrt{y}\\ x&=y\\ x-y&=0\\ \end{align*} Does this imply $\sqrt{x-y}=\sqrt{x}$? I believe this is analogous to what you were asking.

Answer 3

To answer your new questions 3 and 4:

3) If $\beta>\alpha$, the relation still holds. Even more: the chords $PQ$ and $AB$ will be the same (not only equal in distance), because a negative angle will be under the X-axis, and so will be the resulting angle of substracting $\beta$ to $\alpha$.

4) It doesn't matter if $\angle(\alpha-\beta)>\pi$. The relation still holds, because the coordinates of points $P$, $Q$, $A$ and $B$ will still be the same. So, the expressions for chords $PQ$ and $AB$ will be the same, and the rule holds.