Prove that: (6) i) cosa cos(a - b) + sin a sin(a - b) = cos ẞ ii) Cos(a+b).cos(a-β)=cos²a-sin B

Let's go step by step and prove both parts of the question.

### Part (i):
We need to prove:

\[
\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)
\]

#### Using angle difference identities:
We know the following trigonometric identities for angle subtraction:
1. \(\cos(x - y) = \cos(x)\cos(y) + \sin(x)\sin(y)\)
2. \(\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)\)

Now, applying the identity for \(\cos(a - b)\):

\[
\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)
\]

Similarly, for \(\sin(a - b)\):

\[
\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)
\]

Now, substitute these identities into the left-hand side of the given equation:

\[
\cos(a)\cos(a - b) + \sin(a)\sin(a - b)
\]

Substitute \(\cos(a - b)\) and \(\sin(a - b)\) using the identities:

\[
\cos(a)\left[\cos(a)\cos(b) + \sin(a)\sin(b)\right] + \sin(a)\left[\sin(a)\cos(b) - \cos(a)\sin(b)\right]
\]

Now, expand this expression:

\[
= \cos(a)\cos(a)\cos(b) + \cos(a)\sin(a)\sin(b) + \sin(a)\sin(a)\cos(b) - \sin(a)\cos(a)\sin(b)
\]

Simplify the terms:

\[
= \cos^2(a)\cos(b) + \sin^2(a)\cos(b)
\]

Factor out \(\cos(b)\):

\[
= \left(\cos^2(a) + \sin^2(a)\right)\cos(b)
\]

We know from the Pythagorean identity that \(\cos^2(a) + \sin^2(a) = 1\). Therefore, the expression becomes:

\[
= 1 \cdot \cos(b)
\]

Hence, the left-hand side simplifies to:

\[
\cos(b)
\]

This proves that:

\[
\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)
\]

### Part (ii):
We need to prove:

\[
\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)
\]

#### Using angle sum and difference identities:
We know the following identities for angle addition and subtraction:
1. \(\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)\)
2. \(\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)\)

Now, multiply these two expressions:

\[
\cos(a + b)\cos(a - b) = \left(\cos(a)\cos(b) - \sin(a)\sin(b)\right) \left(\cos(a)\cos(b) + \sin(a)\sin(b)\right)
\]

This is in the form of \((x - y)(x + y)\), which simplifies to \(x^2 - y^2\). So, we get:

\[
= \cos^2(a)\cos^2(b) - \sin^2(a)\sin^2(b)
\]

Now, if \(\cos^2(b) = 1 - \sin^2(b)\), substitute this into the equation:

\[
= \cos^2(a)(1 - \sin^2(b)) - \sin^2(a)\sin^2(b)
\]

Expand the expression:

\[
= \cos^2(a) - \cos^2(a)\sin^2(b) - \sin^2(a)\sin^2(b)
\]

Thus, the equation simplifies to:

\[
= \cos^2(a) - \sin^2(b)
\]

Hence, we have proved that:

\[
\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)
\]

---

### Summary:

1. For part (i), we showed that \(\cos(a)\cos(a - b) + \sin(a)\sin(a - b) = \cos(b)\).
2. For part (ii), we demonstrated that \(\cos(a + b)\cos(a - b) = \cos^2(a) - \sin^2(b)\).

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Do you have any questions or would you like more details on any step?

Here are 5 related questions to expand on the topic:
1. What are the other trigonometric identities that can be derived from the angle sum and difference identities?
2. How can the identity for \(\sin(a+b)\) be used in similar problems?
3. What is the relationship between the cosine of complementary angles?
4. How can the identity for \(\cos(a + b)\cos(a - b)\) be applied in physics, particularly in wave interference?
5. Can these identities be extended to hyperbolic trigonometric functions?

**Tip:** Always try to use known identities in trigonometric proofs. Breaking down complex expressions using simple identities like the Pythagorean identity can simplify the process.

Step-by-step proof of two trigonometric identities involving cosine and sine using angle sum and difference formulas. Learn how to apply fundamental trigonometric identities like the Pythagorean identity in proving cos(a + b)cos(a - b) = cos²(a) - sin²(b). Suitable for Grades 10-12.

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