If 7 sin4 θ + 9 cos4 θ + 42 sin2 θ = 16, 0 < θ < \(\frac{\pi}{2}\), then what is tan θ equal to ?
Solving the Trigonometric Equation to Find tan θ
We are given the equation \(7 \sin^4 \theta + 9 \cos^4 \theta + 42 \sin^2 \theta = 16\) and the condition \(0 < \theta < \frac{\pi}{2}\). Our goal is to find the value of \(\tan \theta\). The given condition \(0 < \theta < \frac{\pi}{2}\) tells us that \(\theta\) is in the first quadrant, where \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\) are all positive.
Step 1: Express the equation in terms of a single trigonometric function
We can use the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), which implies \(\cos^2 \theta = 1 - \sin^2 \theta\). We can substitute this into the given equation to express everything in terms of \(\sin \theta\).
The given equation is:
\[7 \sin^4 \theta + 9 \cos^4 \theta + 42 \sin^2 \theta = 16\]Substitute \(\cos^4 \theta = (\cos^2 \theta)^2 = (1 - \sin^2 \theta)^2\):
\[7 \sin^4 \theta + 9 (1 - \sin^2 \theta)^2 + 42 \sin^2 \theta = 16\]Expand the term \((1 - \sin^2 \theta)^2\):
\[(1 - \sin^2 \theta)^2 = 1 - 2 \sin^2 \theta + \sin^4 \theta\]Substitute this back into the equation:
\[7 \sin^4 \theta + 9 (1 - 2 \sin^2 \theta + \sin^4 \theta) + 42 \sin^2 \theta = 16\]Distribute the 9:
\[7 \sin^4 \theta + 9 - 18 \sin^2 \theta + 9 \sin^4 \theta + 42 \sin^2 \theta = 16\]Step 2: Simplify the equation and form a quadratic equation
Combine like terms (\(\sin^4 \theta\) terms and \(\sin^2 \theta\) terms):
\[(7 \sin^4 \theta + 9 \sin^4 \theta) + (-18 \sin^2 \theta + 42 \sin^2 \theta) + 9 = 16\] \[16 \sin^4 \theta + 24 \sin^2 \theta + 9 = 16\]Move the constant term to one side to set the equation to zero:
\[16 \sin^4 \theta + 24 \sin^2 \theta + 9 - 16 = 0\] \[16 \sin^4 \theta + 24 \sin^2 \theta - 7 = 0\]This equation looks like a quadratic equation if we consider \(\sin^2 \theta\) as the variable. Let \(x = \sin^2 \theta\). Since \(0 < \theta < \frac{\pi}{2}\), we know that \(0 < \sin \theta < 1\), which means \(0 < \sin^2 \theta < 1\). So, \(x\) must satisfy \(0 < x < 1\).
Substituting \(x = \sin^2 \theta\) into the equation, we get:
\[16 x^2 + 24 x - 7 = 0\]Step 3: Solve the quadratic equation for \(x = \sin^2 \theta\)
We can solve this quadratic equation by factoring or using the quadratic formula. Let's try factoring. We look for two numbers that multiply to \(16 \times (-7) = -112\) and add up to 24. The numbers 28 and -4 satisfy these conditions (\(28 \times -4 = -112\) and \(28 + (-4) = 24\)).
Rewrite the middle term using 28x and -4x:
\[16 x^2 + 28 x - 4 x - 7 = 0\]Group terms and factor by grouping:
\[4x (4x + 7) - 1 (4x + 7) = 0\] \[(4x - 1)(4x + 7) = 0\]This gives two possible solutions for \(x\):
- \(4x - 1 = 0 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}{4}\)
- \(4x + 7 = 0 \Rightarrow 4x = -7 \Rightarrow x = -\frac{7}{4}\)
Recall that \(x = \sin^2 \theta\) and \(0 < x < 1\). The solution \(x = -\frac{7}{4}\) is not valid because \(\sin^2 \theta\) cannot be negative.
Therefore, the only valid solution is \(x = \frac{1}{4}\).
\[\sin^2 \theta = \frac{1}{4}\]Step 4: Find the value of \(\cos^2 \theta\) and \(\tan^2 \theta\)
Now that we have \(\sin^2 \theta\), we can find \(\cos^2 \theta\) using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):
\[\cos^2 \theta = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}\]Next, we can find \(\tan^2 \theta\) using the identity \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\):
\[\tan^2 \theta = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}\]Step 5: Find the value of \(\tan \theta\)
We have \(\tan^2 \theta = \frac{1}{3}\). To find \(\tan \theta\), we take the square root:
\[\tan \theta = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}\]The problem states that \(0 < \theta < \frac{\pi}{2}\). In the first quadrant (\(0 < \theta < \frac{\pi}{2}\)), the tangent function is positive.
Therefore, we take the positive root:
\[\tan \theta = \frac{1}{\sqrt{3}}\]Conclusion
By simplifying the given trigonometric equation using identities and solving the resulting quadratic equation for \(\sin^2 \theta\), we found \(\sin^2 \theta = \frac{1}{4}\). From this, we calculated \(\cos^2 \theta = \frac{3}{4}\) and finally \(\tan^2 \theta = \frac{1}{3}\). Since \(0 < \theta < \frac{\pi}{2}\), \(\tan \theta\) must be positive, giving us \(\tan \theta = \frac{1}{\sqrt{3}}\).
| Step | Calculation/Reasoning | Result |
|---|---|---|
| 1 | Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) | \(7 \sin^4 \theta + 9(1-\sin^2 \theta)^2 + 42 \sin^2 \theta = 16\) |
| 2 | Expand and simplify | \(16 \sin^4 \theta + 24 \sin^2 \theta - 7 = 0\) |
| 3 | Let \(x = \sin^2 \theta\), solve \(16x^2 + 24x - 7 = 0\) | \(x = \frac{1}{4}\) or \(x = -\frac{7}{4}\) |
| 4 | Select valid solution for \(\sin^2 \theta\) | \(\sin^2 \theta = \frac{1}{4}\) (since \(0 < \sin^2 \theta < 1\)) |
| 5 | Calculate \(\cos^2 \theta = 1 - \sin^2 \theta\) | \(\cos^2 \theta = \frac{3}{4}\) |
| 6 | Calculate \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\) | \(\tan^2 \theta = \frac{1/4}{3/4} = \frac{1}{3}\) |
| 7 | Find \(\tan \theta\) (positive root for \(0 < \theta < \frac{\pi}{2}\)) | \(\tan \theta = \frac{1}{\sqrt{3}}\) |
Revision Table: Key Concepts
This problem utilized several key trigonometric concepts and algebraic techniques.
- Trigonometric Identities: The core identity used was \(\sin^2 \theta + \cos^2 \theta = 1\), allowing us to convert between \(\sin^2 \theta\) and \(\cos^2 \theta\). The definition \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) was also used, specifically \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).
- Solving Quadratic Equations: The problem was reduced to solving a quadratic equation in terms of \(\sin^2 \theta\). Factoring was used in this solution, but the quadratic formula could also be applied.
- Quadrant Analysis: The condition \(0 < \theta < \frac{\pi}{2}\) was crucial for determining the correct sign of \(\tan \theta\) after finding \(\tan^2 \theta\). In the first quadrant, all trigonometric functions are positive.
Additional Information: Solving Trigonometric Equations
Solving trigonometric equations often involves transforming the equation using identities so it contains only one trigonometric function or a factorable form. Here are some common strategies:
- Using Identities: Replace terms using fundamental identities like \(\sin^2 \theta + \cos^2 \theta = 1\), \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), etc., to simplify or unify the equation.
- Factoring: If the equation can be written as a product of factors equal to zero, set each factor to zero and solve the simpler equations.
- Quadratic Form: Recognize equations that are quadratic in form (e.g., \(a \sin^2 \theta + b \sin \theta + c = 0\)) and solve for the trigonometric function (like \(\sin \theta\)) using factoring or the quadratic formula.
- Domain and Range: Always consider the possible values of trigonometric functions (e.g., \(-1 \le \sin \theta \le 1\), \(-1 \le \cos \theta \le 1\)). Solutions outside these ranges are extraneous.
- General Solutions vs. Specific Solutions: Pay attention to the specified interval for \(\theta\). If no interval is given, you usually need to provide general solutions using the periodicity of the trigonometric functions. If an interval like \(0 < \theta < \frac{\pi}{2}\) is given, find the specific solutions within that range.
In this specific problem, recognizing the equation as quadratic in \(\sin^2 \theta\) was a key step after applying the identity to express everything in terms of sine.
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