Solving Trigonometric Equations: Finding sinθ cosθ

We are given a trigonometric equation involving powers of sinθ and cosθ and asked to find the value of their product, sinθ cosθ.

The given equation is:

\(\sin^6 \theta + \cos^6 \theta = \frac{1}{3}\)

where \(0° < \theta < 90°\).

Using Trigonometric Identities to Simplify

To simplify the expression \(\sin^6 \theta + \cos^6 \theta\), we can view it as a sum of cubes. Let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\). Then the expression is \(a^3 + b^3\). We use the algebraic identity \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).

Applying this identity:

\(\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 = (\sin^2 \theta + \cos^2 \theta)((\sin^2 \theta)^2 - (\sin^2 \theta)(\cos^2 \theta) + (\cos^2 \theta)^2)\)

From the fundamental Pythagorean identity, we know that \(\sin^2 \theta + \cos^2 \theta = 1\). Substituting this into the equation:

\(\sin^6 \theta + \cos^6 \theta = (1)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta)\)

\(\sin^6 \theta + \cos^6 \theta = \sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta\)

Further Simplification

Now let's simplify the term \(\sin^4 \theta + \cos^4 \theta\). We can use the identity for the square of a sum: \((a+b)^2 = a^2 + 2ab + b^2\). Let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\).

\((\sin^2 \theta + \cos^2 \theta)^2 = (\sin^2 \theta)^2 + 2(\sin^2 \theta)(\cos^2 \theta) + (\cos^2 \theta)^2\)

\(1^2 = \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta + \cos^4 \theta\)

\(1 = \sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta\)

Rearranging this to isolate \(\sin^4 \theta + \cos^4 \theta\):

\(\sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta\)

Substituting Back and Solving

Substitute this expression for \(\sin^4 \theta + \cos^4 \theta\) back into the simplified equation for \(\sin^6 \theta + \cos^6 \theta\):

\(\sin^6 \theta + \cos^6 \theta = (\sin^4 \theta + \cos^4 \theta) - \sin^2 \theta \cos^2 \theta\)

\(\sin^6 \theta + \cos^6 \theta = (1 - 2 \sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta\)

\(\sin^6 \theta + \cos^6 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta\)

We are given that \(\sin^6 \theta + \cos^6 \theta = \frac{1}{3}\). So, we have:

\(1 - 3 \sin^2 \theta \cos^2 \theta = \frac{1}{3}\)

Now, we solve this equation for \(\sin^2 \theta \cos^2 \theta\):

\(1 - \frac{1}{3} = 3 \sin^2 \theta \cos^2 \theta\)

\(\frac{3 - 1}{3} = 3 \sin^2 \theta \cos^2 \theta\)

\(\frac{2}{3} = 3 \sin^2 \theta \cos^2 \theta\)

\(\sin^2 \theta \cos^2 \theta = \frac{2}{3} \cdot \frac{1}{3}\)

\(\sin^2 \theta \cos^2 \theta = \frac{2}{9}\)

To find \(\sin \theta \cos \theta\), we take the square root of both sides:

\(\sqrt{(\sin \theta \cos \theta)^2} = \sqrt{\frac{2}{9}}\)

\(\sin \theta \cos \theta = \pm \frac{\sqrt{2}}{\sqrt{9}}\)

\(\sin \theta \cos \theta = \pm \frac{\sqrt{2}}{3}\)

Considering the Given Range of θ

The problem specifies that \(0° < \theta < 90°\). This range corresponds to the first quadrant of the unit circle. In the first quadrant, the values of both \(\sin \theta\) and \(\cos \theta\) are positive.

Since \(\sin \theta > 0\) and \(\cos \theta > 0\) for \(0° < \theta < 90°\), their product \(\sin \theta \cos \theta\) must also be positive.

Therefore, we select the positive value:

\(\sin \theta \cos \theta = \frac{\sqrt{2}}{3}\)

Summary of Steps

Revision Table: Useful Identities in Trigonometry

Mastering trigonometric identities is crucial for solving complex problems like this. Here's a quick list of identities used or related:

• Pythagorean Identity: \(\sin^2 x + \cos^2 x = 1\)
• Sum of Cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)
• Square of a Sum: \((a+b)^2 = a^2 + 2ab + b^2\)
• Double Angle Identity for Sine: \(\sin(2\theta) = 2 \sin \theta \cos \theta\) (This could also be used after finding \(\sin \theta \cos \theta\))

Additional Information: Quadrants and Signs of Trigonometric Functions

The quadrant in which the angle \(\theta\) lies determines the signs of \(\sin \theta\), \(\cos \theta\), and other trigonometric functions. This problem specifies \(0° < \theta < 90°\), which is the first quadrant.

Understanding the signs in each quadrant is vital:

In our case, \(0° < \theta < 90°\) places \(\theta\) in Quadrant I, ensuring that \(\sin \theta\) and \(\cos \theta\) are both positive, and thus their product \(\sin \theta \cos \theta\) is also positive.