Question

1. derive: $\cos(u - v) = \cos u \cos v + \sin u \sin v$ i have included some hints below: recall: $x^2 + y^2 = 1$

Explanation:

Step1: Define Points on Unit Circle

Let’s consider the unit circle \(x^{2}+y^{2} = 1\). Let angle \(u\) correspond to the point \(P_1=(\cos u,\sin u)\) and angle \(v\) correspond to the point \(P_2 = (\cos v,\sin v)\) on the unit circle. The angle between the two radii (from the origin to \(P_1\) and \(P_2\)) is \(u - v\).

Step2: Use Distance Formula

The distance between \(P_1\) and \(P_2\) can be calculated using the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). Substituting \(x_1=\cos u,y_1 = \sin u,x_2=\cos v,y_2=\sin v\), we get:
\[
d=\sqrt{(\cos v-\cos u)^{2}+(\sin v - \sin u)^{2}}
\]
Expanding the squares inside the square - root:
\[
d=\sqrt{\cos^{2}v - 2\cos u\cos v+\cos^{2}u+\sin^{2}v - 2\sin u\sin v+\sin^{2}u}
\]
Using the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\) for \(\theta = u\) and \(\theta=v\) (i.e., \(\cos^{2}u+\sin^{2}u = 1\) and \(\cos^{2}v+\sin^{2}v = 1\)):
\[
d=\sqrt{(\cos^{2}u+\sin^{2}u)+(\cos^{2}v+\sin^{2}v)-2(\cos u\cos v+\sin u\sin v)}
\]
\[
d=\sqrt{1 + 1-2(\cos u\cos v+\sin u\sin v)}=\sqrt{2-2(\cos u\cos v+\sin u\sin v)}
\]

Step3: Distance Using Central Angle

We can also calculate the distance between \(P_1\) and \(P_2\) using the law of cosines. For a triangle with two sides of length \(r = 1\) (since it's a unit circle) and the included angle \(u - v\), the distance \(d\) is given by:
\[
d^{2}=1^{2}+1^{2}-2\times1\times1\times\cos(u - v)
\]
\[
d^{2}=2-2\cos(u - v)
\]
Taking the square - root of both sides (since distance is non - negative), we have \(d=\sqrt{2 - 2\cos(u - v)}\)

Step4: Equate the Two Expressions for Distance

Since both expressions represent the distance between \(P_1\) and \(P_2\), we can equate them:
\[
\sqrt{2-2(\cos u\cos v+\sin u\sin v)}=\sqrt{2 - 2\cos(u - v)}
\]
Squaring both sides to eliminate the square - roots:
\[
2-2(\cos u\cos v+\sin u\sin v)=2 - 2\cos(u - v)
\]
Subtract 2 from both sides:
\[
-2(\cos u\cos v+\sin u\sin v)=- 2\cos(u - v)
\]
Divide both sides by \(- 2\):
\[
\cos u\cos v+\sin u\sin v=\cos(u - v)
\]

Answer:

The derivation of \(\cos(u - v)=\cos u\cos v+\sin u\sin v\) is completed as shown above.