Question:
Let \(\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots\) up to 40 terms. If \((\tan\beta)^{\pi/1020}\) is a root of the equation \(x^2 + x - 2 = 0\), \(\beta \in (0, \pi/2)\), then \(\sin^2\beta + 3\cos^2\beta\) is equal to:
Let \(\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots\) up to 40 terms. If \((\tan\beta)^{\pi/1020}\) is a root of the equation \(x^2 + x - 2 = 0\), \(\beta \in (0, \pi/2)\), then \(\sin^2\beta + 3\cos^2\beta\) is equal to:
Updated On: Apr 13, 2026
- \(2 \)
- \(\frac{7}{4} \)
- \(\frac{5}{2} \)
- \(\frac{3}{2} \)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This problem requires solving a quadratic equation to find the value of $\tan \beta$, and then applying trigonometric identities to find the final expression value.
Step 2: Key Formula or Approach:
1. Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
2. Identity: $\sin^2 \theta + \cos^2 \theta = 1$.
Step 3: Detailed Explanation:
First, solve the quadratic equation: \[ x^2 + x - 2 = 0 \] \[ (x + 2)(x - 1) = 0 \] The roots are $x = 1$ and $x = -2$. Since $\beta \in (0, \pi/2)$, $\tan \beta$ must be positive. Therefore: \[ (\tan\beta)^{\pi/1020} = 1 \implies \tan \beta = 1 \] This gives $\beta = \pi/4$ (or $45^\circ$). Now, substitute $\beta = \pi/4$ into the expression: \[ \sin^2(45^\circ) + 3\cos^2(45^\circ) \] \[ = \left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2 \] \[ = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2 \]
Step 4: Final Answer:
The value of $\sin^2\beta + 3\cos^2\beta$ is 2.
This problem requires solving a quadratic equation to find the value of $\tan \beta$, and then applying trigonometric identities to find the final expression value.
Step 2: Key Formula or Approach:
1. Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
2. Identity: $\sin^2 \theta + \cos^2 \theta = 1$.
Step 3: Detailed Explanation:
First, solve the quadratic equation: \[ x^2 + x - 2 = 0 \] \[ (x + 2)(x - 1) = 0 \] The roots are $x = 1$ and $x = -2$. Since $\beta \in (0, \pi/2)$, $\tan \beta$ must be positive. Therefore: \[ (\tan\beta)^{\pi/1020} = 1 \implies \tan \beta = 1 \] This gives $\beta = \pi/4$ (or $45^\circ$). Now, substitute $\beta = \pi/4$ into the expression: \[ \sin^2(45^\circ) + 3\cos^2(45^\circ) \] \[ = \left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2 \] \[ = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2 \]
Step 4: Final Answer:
The value of $\sin^2\beta + 3\cos^2\beta$ is 2.
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