Q
Numerical
Compound Angles
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha – \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in N$, then $r + s$ is equal to ________ .
1. Compound Angle Formulae
The formula $\tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}$ is a staple of trigonometric algebra. It allows us to determine the tangent of a sum if we know the tangents of the individual parts. In advanced JEE problems, the “parts” are often given as $( \alpha + \beta)$ and $(\alpha – \beta)$, and the target is $2\alpha$ (the sum) or $2\beta$ (the difference). Mastering these manipulations is crucial for speed.
2. Quadrant Analysis for Inverse Trig
When calculating $\sin x$ or $\cos x$ from each other using $\sin^2 x + \cos^2 x = 1$, the sign of the square root depends entirely on the quadrant. In this problem, the bounds $0 < \alpha < \pi/3$ and $0 < \beta < \pi/4$ determine the possible range for $\alpha \pm \beta$. Since $\cos(\alpha+\beta) < 0$, it immediately flags that the angle must be obtuse, ensuring we choose the positive root for $\sin(\alpha+\beta)$ as it lies in the 2nd quadrant.
3. Radical Simplification
Expressions like $\sqrt{605}$ often appear in JEE to test mental calculation. Recognizing that $605 = 121 \times 5$ allows for the simplification to $11\sqrt{5}$. Similarly, $\sqrt{55}$ is $\sqrt{11} \times \sqrt{5}$. Factoring out these common radicals (like $\sqrt{11}$ in the denominator) is the key to matching the final result to the given algebraic form.
4. Numerical Comparison Technique
Numerical type questions often provide an expression with variables like $r, s$. Once you simplify your mathematical result into a form that looks identical to the given structure, you can equate the coefficients. This is a common pattern in the Integer/Numerical section where the final answer is a simple arithmetic sum of these discovered coefficients.
1
Why did we use tan(2α) = tan((α+β)+(α-β))?
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Because we were given values for functions of (α+β) and (α-β). Summing these two quantities eliminates β, leaving exactly 2α.
2
How do we know sin(α+β) is positive?
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The sum α+β is less than π, and sine is positive in both the 1st and 2nd quadrants.
3
How did √605 become 11√5?
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605 can be factored as 5 × 121. Since 121 is the square of 11, the square root is 11√5.
Yes, technically α-β can be negative if β > α. However, sin(α-β) = 3/8 is positive, which tells us α-β must be positive in this specific case.
5
Why is tan(α+β) = -3√11?
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tan = sin/cos. Here, (3√11/10) / (-1/10) = -3√11. The negative sign comes from the 2nd quadrant location.
6
What is the range of tan x in the 2nd quadrant?
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In the 2nd quadrant, tangent values range from -∞ to 0.
7
How do we solve for tan 2β?
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To find tan 2β, we would use the subtraction identity: tan((α+β) – (α-β)).
8
Are r and s always integers?
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The problem statement explicitly says $r, s \in N$, meaning they must be natural numbers (positive integers).
9
What if sin(α-β) was negative?
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Then α-β would be in the 4th quadrant, making tan(α-β) negative, which would change the sum in the formula.
10
Why simplify radicals before the final step?
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It makes it much easier to spot common factors like √11 or √5, which are needed to match the target expression format.