Question

The equation \(\sin ^{6} x+\cos ^{6} x=a^{2}\) has real solutions if a belongs to (a) \(\left[\frac{-1}{2}, \frac{1}{2}\right]\) (b) \(\left[-1, \frac{-1}{2}\right]\) or \(\left[\frac{1}{2}, 1\right]\) (c) \([-1,1]\) (d) \(\left[\frac{-1}{2}, 1\right]\)

Short answer

Answer: The range of values for \(a\) is \(\left[\frac{1}{2}, 1\right]\).

Step-by-step solution

Analyzing the given equation

The given equation is \(\sin^6 x + \cos^6 x = a^2\). To find an alternative expression, first, notice that \(\sin^2 x + \cos^2 x = 1\). Let's square this equation and manipulate it to obtain an expression containing \(\sin^6 x + \cos^6 x\).

Squaring the equation and manipulation

We can start by squaring the equation \(\sin^2 x + \cos^2 x = 1\): $$ (\sin^2 x + \cos^2 x)^2 = 1^2 $$ Expanding the left side of the equation, we get: $$ \sin^4 x + 2\sin^2 x\cos^2 x + \cos^4 x = 1 $$ Now, we need to obtain an expression containing \(\sin^6 x + \cos^6 x\). To do this, let's take the expression above and square one more time.

Squaring once again and simplifying

Again squaring the equation, we get: $$ \left(\sin^4 x + 2\sin^2 x\cos^2 x + \cos^4 x\right)^2 = 1^2 $$ Expanding and simplifying the left side of the equation: $$ (\sin^4 x + \cos^4 x + 2\sin^2 x\cos^2 x)^2 = (1 - 2\sin^2 x\cos^2 x)^2 = 1 - 4\sin^4 x\cos^4 x + 4\sin^6 x\cos^6 x $$ Now we can rewrite the expression \(\sin^6 x + \cos^6 x\), using the relation: $$ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x $$

Solving for \(\sin^6 x + \cos^6 x\)

We have the expression: $$ \sin^6 x + \cos^6 x = (\sin^4 x + \cos^4 x - 2\sin^2 x\cos^2 x)^2 = (1 - 4\sin^2 x\cos^2 x + 2\sin^4 x\cos^4 x)^2 $$ Now we can find the range of values for \(a\).

Finding the range of values of \(a\)

We know that \(\sin^2 x\) and \(\cos^2 x\) are both non-negative and their sum is 1. So, if we let \(t = \sin^2 x\cos^2 x\), then we have \(t \geq 0\) and \(t\leq \frac{1}{4}\) (by the arithmetic mean-geometric mean inequality). Replacing \(t\) in the equation, we get: $$ a^2 = (1 - 4t + 2t^2)^2 $$ This is a quadratic expression in terms of \(t\). The minimum value of this expression can be found by analyzing the corresponding parabola. Since the coefficient of \(t^2\) is positive, the parabola has a minimum. To find this minimum, we take the first derivative of the expression with respect to \(t\) and set it to zero. $$ \frac{d(1 - 4t + 2t^2)^2}{dt} = 0 $$ Solving for \(t\), we get: $$ t = \frac{1}{2} $$ Substituting this value back into the expression for \(a^2\), we get the minimum value of \(a^2\): $$ a^2 = \left(1 - 4\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right)^2\right)^2 = \frac{1}{4} $$ Since \(t\leq \frac{1}{4}\), the maximum value of the expression is when \(t = 0\): $$ a^2 = (1 - 4(0) + 2(0)^2)^2 = 1^2 = 1 $$ Therefore, the range of values for \(a\) such that the equation \(\sin^6 x + \cos^6 x = a^2\) has real solutions is \(a \in \left[\frac{1}{2}, 1\right]\). This corresponds to option (d).

Key concepts

Sine and Cosine

Understanding the basic properties of sine and cosine is crucial in solving trigonometric equations. These functions relate to angles and triangles in geometry, but they also connect to unit circles in algebra. In every angle,

• The sine function (\(sin(x)\)) gives the vertical coordinate of a point on the unit circle.
• The cosine function (\(cos(x)\)) gives the horizontal coordinate of that point.

Together, sine and cosine play vital roles in forming identities like \(sin^2 x + cos^2 x = 1\), which expresses a fundamental property of these trigonometric functions. By squaring and manipulating this equation, one can derive more complex equations used to explore real solutions.

Quadratic Expressions

Quadratic expressions come into play in this problem as \(a^2 = (1 - 4t + 2t^2)^2\). These expressions take the form \(ax^2 + bx + c\), and they can represent parabolas when graphed.
A key technique in quadratic problems is to understand how the expression behaves, particularly noting:

• Whether the parabola opens upwards or downwards (determined by the sign of the leading coefficient).
• The vertex of the parabola, which can show minimum or maximum values.

In our case, finding the minimum value of a quadratic helps identify the range for \(a^2\), guiding to the real solutions of the equation.

Real Solutions

Finding real solutions means identifying values that satisfy the equation in the real number system. For the equation \(sin^6 x + cos^6 x = a^2\), we need solutions in real numbers for it to hold true.
Real solutions are usually found by analyzing:

• The nature of the expression (e.g., whether it forms a quadratic equation).
• The feasible values for trigonometric functions, which are always between -1 and 1.
• The specific boundaries or values where an equation reaches its minimum or maximum.

In this exercise, the values are determined by identifying where the derived quadratic expression has its critical points within permissible limits.

AM-GM Inequality

The AM-GM inequality is a powerful tool in mathematics, often used to find extremum values. It states that for any non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean. In the context of our problem:

• The inequality helps estimate the maximum and minimum of expressions like \(t = sin^2 x cos^2 x\).
• It guides us to understand that \(t\) achieves its upper bound at \(\frac{1}{4}\).

By applying this inequality, we ensure the solutions fall within expected bounds, allowing us to identify the proper domain for \(a\) that yields real solutions.