Q.If $\sin x + \cos x = a$, then

The key idea here is applying algebraic identities to trigonometric expressions, leveraging the fundamental identity $\sin^2 x + \cos^2 x = 1$.

(i) To find $\sin^6 x + \cos^6 x$:

1. Square the given equation: $(\sin x + \cos x)^2 = a^2 \implies \sin^2 x + \cos^2 x + 2 \sin x \cos x = a^2$.
2. Using $\sin^2 x + \cos^2 x = 1$, we get $1 + 2 \sin x \cos x = a^2$, which means $2 \sin x \cos x = a^2 - 1$.
3. Rewrite $\sin^6 x + \cos^6 x$ as $(\sin^2 x)^3 + (\cos^2 x)^3$. Using the identity $A^3 + B^3 = (A+B)(A^2 - AB + B^2)$, this becomes $(\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2)$.
4. Substitute $\sin^2 x + \cos^2 x = 1$ and $(\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$: $1 \cdot (1^2 - 2 \sin^2 x \cos^2 x - \sin^2 x \cos^2 x) = 1 - 3 \sin^2 x \cos^2 x$.
5. Substitute $2 \sin x \cos x = a^2 - 1$: …

The core idea behind solving this problem is to leverage fundamental algebraic identities and the primary trigonometric identity $\sin^2 x + \cos^2 x = 1$. When we are given a sum like $\sin x + \cos x = a$, squaring this expression is often the first step, as it introduces the term $2 \sin x \cos x$, which is crucial for simplifying many other trigonometric expressions.

For part (i), $\sin^6 x + \cos^6 x$, we recognize this as a sum of cubes, specifically $(\sin^2 x)^3 + (\cos^2 x)^3$. This allows us to use the algebraic identity for $A^3 + B^3$, which simplifies significantly because $A+B = \sin^2 x + \cos^2 x = 1$. The remaining terms will involve $\sin x \cos x$, which we can express in terms of $a$.

For part (ii), $|\sin x - \cos x|$, we want to find the absolute value of a difference. It's often easier to work with the square of the difference, $(\sin x - \cos x)^2$, because it also simplifies using $\sin^2 x + \cos^2 x = 1$ and involves $\sin x \cos x$. Once we have the square, taking the square root gives us the absolute value.

Let's proceed with the steps.

Given: $\sin x + \cos x = a$

1. Find $\sin x \cos x$ in terms of $a$. This is a common first step when $\sin x + \cos x$ or $\sin x - \cos x$ is given. Square both sides of the given equation:

Expand the left side:

Using the identity $\sin^2 x + \cos^2 x = 1$:

Now, isolate $\sin x \cos x$:

This expression for $\sin x \cos x$ will be used in both parts of the problem.

> [!WARNING]
> Remember that for $\sin x + \cos x = a$ to be valid, $a$ must be in the range $[-\sqrt{2}, \sqrt{2}]$. This is because $\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$, and the maximum value of $\sin(\theta)$ is $1$ and the minimum is $-1$. Thus, $-\sqrt{2} \le a \le \sqrt{2}$. This implies $a^2 \le 2$.

(i) Find $\sin^6 x + \cos^6 x$

2. Recognize the expression as a sum of cubes.

   We can write $\sin^6 x + \cos^6 x$ as $(\sin^2 x)^3 + (\cos^2 x)^3$.

   This fits the algebraic identity $A^3 + B^3 = (A+B)^3 - 3AB(A+B)$.

   $A^3 + B^3 = (A+B)^3 - 3AB(A+B)$

3. Apply the identity with $A = \sin^2 x$ and $B = \cos^2 x$.

Substitute $\sin^2 x + \cos^2 x = 1$:

4. Substitute the value of $\sin x \cos x$ from Step 1. We found $\sin x \cos x = \frac{a^2 - 1}{2}$. Substitute this into the expression:

Expand $(a^2 - 1)^2$:

(ii) Find $|\sin x - \cos x|$

5. Consider the square of the expression. Let $Y = \sin x - \cos x$. We want to find $|Y|$. It's easier to find $Y^2$ first: …