In trigonometric simplification, consider the expression (sin θ)^6 + (cos θ)^6 + 3 * (sin θ)^2 * (cos θ)^2 − 1. Simplify this expression to a single constant value that is independent of θ.

Introduction / Context:
This problem tests your ability to simplify higher power trigonometric expressions using algebraic identities and the fundamental relation between sine and cosine. By combining sixth powers of sin θ and cos θ with a mixed product term and a constant, the question encourages you to recognise patterns similar to algebraic identities involving cubes and the fact that sin^2 θ + cos^2 θ = 1. Such expressions often reduce to simple constants if handled carefully, which is a common theme in aptitude and exam questions.

Given Data / Assumptions:

• The expression to simplify is (sin θ)^6 + (cos θ)^6 + 3 * (sin θ)^2 * (cos θ)^2 − 1.
• θ is any real angle for which sine and cosine are defined.
• The standard identity sin^2 θ + cos^2 θ = 1 holds.
• We want a simplification that does not depend on θ at all.

Concept / Approach:
The key idea is to treat sin^2 θ and cos^2 θ as algebraic variables. Let s = sin^2 θ and c = cos^2 θ. Then s + c = 1. The expression becomes s^3 + c^3 + 3 s c. The sum s^3 + c^3 can be simplified using the algebraic identity s^3 + c^3 = (s + c)^3 − 3 s c (s + c). Substituting s + c = 1 greatly simplifies this expression. Once that is done, the + 3 s c and −1 terms can be combined, and we will see that everything cancels.

Step-by-Step Solution:
1) Set s = sin^2 θ and c = cos^2 θ. Then s + c = 1 by the Pythagoras identity. 2) Rewrite the expression in terms of s and c: s^3 + c^3 + 3 s c − 1. 3) Use the identity s^3 + c^3 = (s + c)^3 − 3 s c (s + c). 4) Since s + c = 1, we have s^3 + c^3 = 1^3 − 3 s c * 1 = 1 − 3 s c. 5) Substitute this back into the expression: (1 − 3 s c) + 3 s c − 1. 6) Combine like terms: the −3 s c and +3 s c cancel, and 1 − 1 cancels as well. 7) The result is 0, which is independent of θ.

Verification / Alternative check:
You can test this result by picking specific values of θ. For example, let θ = 0°. Then sin θ = 0 and cos θ = 1. The expression becomes 0^6 + 1^6 + 3 * 0^2 * 1^2 − 1 = 0 + 1 + 0 − 1 = 0. For θ = 45°, sin θ = cos θ = √2 / 2, so sin^2 θ = cos^2 θ = 1/2. Then s^3 + c^3 = 2 * (1/2)^3 = 2 * 1/8 = 1/4 and 3 s c = 3 * (1/2) * (1/2) = 3/4. The full expression is 1/4 + 3/4 − 1 = 1 − 1 = 0. These checks support the algebraic simplification.

Why Other Options Are Wrong:
Options b (1), c (2), d (−1), and e (4) would mean that the expression always evaluates to a non zero constant, but direct substitution for simple angles like 0° shows the result is 0. Any non zero choice can be ruled out quickly with such a test. Only option a reproduces the value obtained from both the identity based simplification and numerical checks at sample angles.

Common Pitfalls:
One frequent mistake is to attempt expanding everything directly in terms of sin θ and cos θ without re grouping, which leads to very complicated algebra. Another issue is forgetting to use the identity sin^2 θ + cos^2 θ = 1 to replace s + c with 1. Some learners mis apply the sum of cubes formula, writing s^3 + c^3 as (s + c)^3 without the correction term. Carefully using s and c as temporary variables and applying the correct algebraic identity avoids these errors.

Final Answer:
After simplification, the expression (sin θ)^6 + (cos θ)^6 + 3 * (sin θ)^2 * (cos θ)^2 − 1 is identically equal to 0 for all θ.