Question

If A, B, C, D are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity k, then the value of $4sin\frac{A}{2}+3sin\frac{B}{2}+2sin\frac{C}{2}+sin\frac{D}{2}$ is equal to :

• A. $2\sqrt{1-k}$
• B. $\sqrt{1+k}$
• C. $2\sqrt{k}$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

To solve this problem, we need to first identify the angles A, B, C, D based on the given conditions and then substitute them into the expression.

1. Identify the angles A, B, C, D:
   We are given that A, B, C, D are the smallest positive angles in ascending order of magnitude such that their sines are equal to a positive quantity $k$.
   Let $\sin x = k$. Since $k > 0$, the angles must lie in quadrants where sine is positive (Quadrant I or II).
   Let A be the principal value, so $A = \arcsin k$. Since A is the smallest positive angle, $0 < A < \frac{\pi}{2}$.

   The general solution for $\sin x = k$ is $x = n\pi + (-1)^n A$, where $n$ is an integer.
   The smallest positive angles in ascending order are:

   • For $n=0$: $A = 0\cdot\pi + (-1)^0 A = A$.
   • For $n=1$: $B = 1\cdot\pi + (-1)^1 A = \pi - A$.
   • For $n=2$: $C = 2\cdot\pi + (-1)^2 A = 2\pi + A$.
   • For $n=3$: $D = 3\cdot\pi + (-1)^3 A = 3\pi - A$.

   These angles are indeed in ascending order: $A < \pi - A < 2\pi + A < 3\pi - A$ for $A \in (0, \pi/2)$.

2. Substitute the angles into the expression:
   The expression to evaluate is $E = 4\sin\frac{A}{2}+3\sin\frac{B}{2}+2\sin\frac{C}{2}+sin\frac{D}{2}$.
   Substitute $B = \pi - A$, $C = 2\pi + A$, and $D = 3\pi - A$:
   $E = 4\sin\frac{A}{2} + 3\sin\left(\frac{\pi - A}{2}\right) + 2\sin\left(\frac{2\pi + A}{2}\right) + \sin\left(\frac{3\pi - A}{2}\right)$
   $E = 4\sin\frac{A}{2} + 3\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) + 2\sin\left(\pi + \frac{A}{2}\right) + \sin\left(\frac{3\pi}{2} - \frac{A}{2}\right)$

3. Apply trigonometric identities:
   We use the following identities:

   • $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$
   • $\sin(\pi + \theta) = -\sin\theta$
   • $\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta$

   Applying these identities with $\theta = \frac{A}{2}$:
   $E = 4\sin\frac{A}{2} + 3\cos\frac{A}{2} + 2\left(-\sin\frac{A}{2}\right) + \left(-\cos\frac{A}{2}\right)$
   $E = 4\sin\frac{A}{2} + 3\cos\frac{A}{2} - 2\sin\frac{A}{2} - \cos\frac{A}{2}$

4. Combine like terms:
   $E = (4-2)\sin\frac{A}{2} + (3-1)\cos\frac{A}{2}$
   $E = 2\sin\frac{A}{2} + 2\cos\frac{A}{2}$
   $E = 2\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)$

5. Express in terms of k:
   We know the identity $(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + \sin(2\theta)$.
   Let $\theta = \frac{A}{2}$. Then $2\theta = A$.
   $\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right)^2 = 1 + \sin A$
   Since $0 < A < \frac{\pi}{2}$, it follows that $0 < \frac{A}{2} < \frac{\pi}{4}$. In this range, both $\sin\frac{A}{2}$ and $\cos\frac{A}{2}$ are positive.
   Therefore, $\sin\frac{A}{2} + \cos\frac{A}{2} = \sqrt{1 + \sin A}$.
   Given that $\sin A = k$:
   $\sin\frac{A}{2} + \cos\frac{A}{2} = \sqrt{1 + k}$

6. Final result:
   Substitute this back into the expression for E:
   $E = 2\left(\sin\frac{A}{2} + \cos\frac{A}{2}\right) = 2\sqrt{1 + k}$

Comparing this result with the given options: (A) $2\sqrt{1-k}$ (B) $\sqrt{1+k}$ (C) $2\sqrt{k}$ (D) None of these

Our calculated value is $2\sqrt{1+k}$, which is not exactly matched by any of the options (A), (B), or (C). Option (B) is $\sqrt{1+k}$, which is half of our result. Therefore, the correct option is (D) None of these.