Question

Which derivation correctly uses the cosine sum identity to prove the cosine double angle identity?.

Answer 1

The cosine sum identity is:

`\[ \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \]`

This is the cosine double-angle identity in its first form.

`\[ \cos(2A) = \cos^2(A) - \sin^2(A) \]`

To derive the cosine double-angle identity using the cosine sum identity, we start with the known sum identity for cosine and then apply it to the specific case of a double angle. The cosine sum identity is:

`\[ \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \]`

The cosine double-angle identity we want to prove is one of the following forms:

1.
`\( \cos(2A) = \cos^2(A) - \sin^2(A) \)`

2.
`\( \cos(2A) = 2\cos^2(A) - 1 \)`

3.
`\( \cos(2A) = 1 - 2\sin^2(A) \)`

Let's prove the first form and then show how it can lead to the other two forms.

Step 1: Apply the Cosine Sum Identity to a Double Angle

Let's set A = B in the cosine sum identity:

`\[ \cos(A + A) = \cos(A)\cos(A) - \sin(A)\sin(A) \]`

`\[ \cos(2A) = \cos^2(A) - \sin^2(A) \]`

This is the cosine double-angle identity in its first form.

Step 2: Derive the Other Forms

1. To derive the second form, use the Pythagorean identity
`\( \sin^2(A) = 1 - \cos^2(A) \):`

`\[ \cos(2A) = \cos^2(A) - (1 - \cos^2(A)) \]`

`\[ \cos(2A) = 2\cos^2(A) - 1 \]`

2. To derive the third form, use the Pythagorean identity
`\( \cos^2(A) = 1 - \sin^2(A) \):`

`\[ \cos(2A) = (1 - \sin^2(A)) - \sin^2(A) \]`

`\[ \cos(2A) = 1 - 2\sin^2(A) \]`

Conclusion

Thus, we have derived the cosine double-angle identity from the cosine sum identity and also shown how it can be expressed in multiple forms.