Question

Challenge Problem If \(2 \sin ^{2} \theta+3 \cos ^{2} \theta=3 \sin \theta \cos \theta+1\) with \(\theta\) in quadrant I, find the possible values for \(\cot \theta\).

Short answer

\(cot \theta = 1\)

Step-by-step solution

- State the given equation

The given equation is:\[2 \sin^{2} \theta + 3 \cos^{2} \theta = 3 \sin \theta \cos \theta + 1\]

- Use Pythagorean Identity

Recall the Pythagorean identity: \[\sin^2 \theta + \cos^2 \theta = 1\]This can be rearranged to express \(\cos^2 \theta\):\[\cos^2 \theta = 1 - \sin^2 \theta\]Substitute \(\cos^2 \theta\) into the given equation:

- Substitute \(\cos^2 \theta\)

Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the equation:\[2 \sin^{2} \theta + 3 \(1 - \sin^{2} \theta\) = 3 \sin \theta \cos \theta + 1\]Simplify the equation:\[2 \sin^{2} \theta + 3 - 3 \sin^{2} \theta = 3 \sin \theta \cos \theta + 1\]Combine like terms:\[- \sin^{2} \theta + 3 = 3 \sin \theta \cos \theta + 1\]

- Simplify to get a quadratic equation

Move all terms to one side of the equation:\[- \sin^{2} \theta - 3 \sin \theta \cos \theta + 2 = 0\]Since \(\sin \theta = \sin \theta\) and \(\cos \theta = \cos \theta\), let's use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to simplify the equation:

- Substitute \(\sin 2\theta\)

Recognize that \(2 \sin \theta \cos \theta = \sin 2\theta\). Substitute \(\sin 2\theta\) into the equation:\[- \sin^{2} \theta - \frac{3}{2} \sin 2\theta + 2 = 0\]This represents a quadratic equation in terms of \(\sin \theta\) and \(\sin 2\theta\).

- Solve the quadratic equation

To solve this quadratic equation, we'll isolate \(\sin \theta\) and \(\sin 2\theta\):First, express \( \sin \theta = x\) for simplicity:\[- x^{2} - 1.5 \sin 2\theta + 2 = 0\]Which is equivalent to:\[- x^{2} - 1.5 ( 2x \sqrt{1 - x^{2}} ) + 2 = 0\]Solve this equation for \(x\), which corresponds to \(\sin \theta\). Since \( \theta \) is in quadrant I, \(\cot \theta > 0 \) can be found by finding the value(s) of \( \cot \theta \) contained.

- Calculate \(\cot \theta\)

Having isolated \(\sin \theta\), recall that\[ \cot \theta = \frac { \cos \theta } { \sin \theta} \]Using results from previous steps we can evaluate for \( x \)

Key concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It states that for any angle \(\theta\), the relationship between sine and cosine can be expressed as: \[ \sin^2 \theta + \cos^2 \theta = 1 \] This identity is derived from the Pythagorean theorem, which gives us a way to relate the sine and cosine of an angle. To solve the given equation, we needed to rewrite \( \cos^2 \theta \) in terms of \( \sin^2 \theta \). By rearranging the Pythagorean identity, we can express cosine squared as: \[ \cos^2 \theta = 1 - \sin^2 \theta \] Substituting this into the problem helps simplify the equation, making it easier to solve. This substitution converts our equation to the same variable, allowing for simpler manipulation and solving.

Sine and Cosine

Sine and cosine are the fundamental trigonometric functions. They relate the angles of a right triangle to the lengths of its sides. For an angle \( \theta \) in a right triangle:

• \( \sin \theta = \frac{opposite}{hypotenuse} \)
• \( \cos \theta = \frac{adjacent}{hypotenuse} \)

Understanding these functions is key to solving trigonometric equations. In this problem, we mainly dealt with their squared values, incorporating the Pythagorean identity. Additionally, knowing the relationship between the trigonometric functions helped in simplifying and substituting values to ultimately solve for \( \theta \). Understanding relationships like \( \sin 2\theta = 2 \sin \theta \cos \theta \) was crucial for simplifying the given equation. As \( \theta \) was given to be in the first quadrant, both sine and cosine are positive, making the values we calculate straightforward.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. Standardly written as: \[ ax^2 + bx + c = 0 \] In our trigonometric problem, we transformed the given trigonometric equation into a quadratic form by simplifying and using identities. The final form of our equation was: \[ - \sin^2 \theta - \frac{3}{2} \sin 2\theta + 2 = 0 \] Solving this involved typical methods used for quadratic equations, such as factorization or the quadratic formula. In this context, substitution made the problem more manageable. We simplified our quadratic terms by representing \( \sin \theta = x \). This standard quadratic approach allowed us to isolate the variable and solve the equation. Understanding and applying quadratic solutions are essential in a variety of mathematical contexts, including solving trigonometric equations like the one provided in this exercise.