91Ó°ÊÓ

The Pythagorean Identity is a fundamental concept in trigonometry. It states that for any angle \(x\), \[ \sin^2 x + \cos^2 x = 1 \] This identity is derived from the Pythagorean theorem in geometry, relating the sides of a right-angled triangle. In our original exercise, we utilize this identity to simplify trigonometric expressions.
When solving the equation \(\sin^6 x + \cos^6 x = \frac{7}{16}\), we first rewrite it as \((\sin^2 x)^3 + (\cos^2 x)^3\). This highlights the importance of the Pythagorean Identity, as it allows us to substitute \(\sin^2 x + \cos^2 x\) with 1, and thus, simplify the expression. We see its practical utility in turning a potentially complex trigonometric equation into a more manageable form. Understanding this identity can make solving similar problems much easier.

The Cube Sum Formula is particularly useful when dealing with cube terms in equations. It states that for any two numbers \(a\) and \(b\), \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] We used this formula in our trigonometric exercise to further simplify the expression \((\sin^2 x)^3 + (\cos^2 x)^3\).
By applying the Cube Sum Formula, the expression can be rewritten as \[ (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2x + (\cos^2 x)^2) \] Knowing that \( \sin^2 x + \cos^2 x = 1 \), this further reduces to: \[ 1((1 - 3\sin^2 x \cos^2 x)) \] This simplification is pivotal in managing and solving the original equation. A clear understanding of the cube sum formula aids in transforming potential hurdles into straightforward paths to the solution by approaching cubes strategically.

When dealing with trigonometric expressions, we often encounter quadratic equations. In the context of the trigonometric exercise, we were led to a quadratic equation form by replacing \(\sin^2 x\) with \(1 - \cos^2 x\).
Our equation simplifies to:\[ 3\cos^4 x - 3\cos^2 x + \frac{9}{16} = 0 \] This is a standard form of a quadratic equation in terms of \(\cos^2 x\). Solving quadratic equations involves:

• Rearranging the equation into standard form: \(ax^2 + bx + c = 0\).
• Applying the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In this exercise, we solve for \(\cos^2 x\), and further steps involve taking the square root to find \(\cos x\). The solution involves considering the symmetry of the cosine function to find all possible values for \(x\). Understanding how to handle quadratic equations such as these is essential in trigonometry, as they frequently appear in various problems.