Answer:
See Below.
Step-by-step explanation:
We are given that:
[tex]\displaystyle \cos(A - B) = \cos A\cos B + \sin A \sin B[/tex]
Part 5.2.1
Note that:
[tex]\displaystyle \cos ( A + B) = \cos (A - (-B))[/tex]
Then from the given identity:
[tex]\displaystyle \cos(A - (-B)) = \cos A \cos( -B) + \sin A \sin (-B)[/tex]
Cosine is an even function and sine is an odd function. That is:
[tex]\displaystyle \cos(- \theta) = \cos \theta \text{ and } \sin (-\theta) = -\sin\theta[/tex]
Hence:
[tex]\displaystyle \cos(A + B) = \cos A \cos B - \sin A \sin B[/tex]
Part 5.2.2
We want to verify that:
[tex]\displaystyle \cos^2 (A - B) = 4\cos^2 B \sin^2 B[/tex]
From the identity:
[tex]\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B[/tex]
Since A + B = 90°, A = 90° - B. Hence:
[tex]\displaystyle \cos (A - B) = \cos (90^\circ - B) \cos B + \sin (90^\circ - B) \sin B[/tex]
Sine and cosine are co-functions. That is:
[tex]\displaystyle \sin \theta = \cos (90^\circ - \theta) \text{ and } \cos\theta = \sin(90^\circ - \theta)[/tex]
Hence:
[tex]\displaystyle \cos (A - B) = \sin B\cos B + \sin B \cos B = 2\sin B \cos B[/tex]
And by squaring both sides:
[tex]\displaystyle \cos ^2 (A - B) = 4\sin^2 B\cos^2 B[/tex]
