Question

If $\alpha ,\beta ,\gamma ,\delta $ are the smallest positive angles in ascending order of magnitude which have their sines equal to positive quantity $\lambda $ then the value of $4\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2}$ is
$
  A.{\text{ }}2\sqrt {1 - \lambda } \\
  B.{\text{ }}2\sqrt {1 + \lambda } \\
  C.{\text{ }}2\sqrt \lambda \\
  D.{\text{ }}2\sqrt {\lambda + 2} \\
$

Answer 1

Hint: In this question use the concept that sine is positive in the first and second quadrant and the angles is in ascending order so the order of ascending is $\alpha < \beta < \gamma < \delta $ so, use this concept to reach the solution of the question.

Complete step-by-step answer:

Given condition is
$\alpha ,\beta ,\gamma ,\delta $ are the smallest positive angles in ascending order of magnitude which have their sines equal to positive quantity $\lambda $.
$ \Rightarrow \sin \alpha = \sin \beta = \sin \gamma = \sin \delta = \lambda $………………. (1)
Then we have to find out the value of
$4\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2}$…………………. (2)
Now from equation (1)
$\sin \alpha = \sin \beta $
$ \Rightarrow \sin \beta = \sin \left( {\pi - \alpha } \right)$ (As we know sine is positive in first and second quadrant and the angles is in ascending order, the order of ascending is $\alpha < \beta < \gamma < \delta $).
$ \Rightarrow \beta = \pi - \alpha $
Similarly $\left( {\gamma = 2\pi + \alpha } \right),\left( {\delta = 3\pi - \alpha } \right)$.
Now substitute all these values in equation (2) we have
$ \Rightarrow 4\sin \dfrac{\alpha }{2} + 3\sin \dfrac{{\pi - \alpha }}{2} + 2\sin \dfrac{{2\pi + \alpha }}{2} + \sin \dfrac{{3\pi - \alpha }}{2}$
Now simplify the above equation we have
$ \Rightarrow 4\sin \dfrac{\alpha }{2} + 3\sin \left( {\dfrac{\pi }{2} - \dfrac{\alpha }{2}} \right) + 2\sin \left( {\pi + \dfrac{\alpha }{2}} \right) + \sin \left( {\dfrac{{3\pi }}{2} - \dfrac{\alpha }{2}} \right)$
Now as we know $\left[ {\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta ,\sin \left( {\pi + \theta } \right) = - \sin \theta ,\sin \left( {\dfrac{{3\pi }}{2} - \theta } \right) = - \cos \theta } \right]$ so, substitute these values in above equation we have,
$ \Rightarrow 4\sin \dfrac{\alpha }{2} + 3\cos \dfrac{\alpha }{2} - 2\sin \dfrac{\alpha }{2} - \cos \dfrac{\alpha }{2}$
$ \Rightarrow 2\left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right)$…………………………. (3)
Now as we know ${\left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right)^2} = {\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} + 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}$
And the value of ${\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} = 1,{\text{ & }}2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} = 2\sin \alpha $
$
   \Rightarrow {\left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right)^2} = 1 + \sin \alpha \\
   \Rightarrow \left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right) = \sqrt {1 + \sin \alpha } \\
$
So substitute this value in equation (3) we have,

$ \Rightarrow 4\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2} = 2\left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right) = 2\sqrt {1 + \sin \alpha } $
Now from equation (1) we have $\sin \alpha = \lambda $
\[ \Rightarrow 4\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2} = 2\sqrt {1 + \lambda } \]
So, this is the required answer.
Hence, option (B) is correct.

Note: In such types of questions angles is in ascending order as $\alpha < \beta < \gamma < \delta $and it is also given that $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = \lambda $ so, according to property of sine which is stated above, $\alpha $ lies in first quadrant, $\beta $ lies in second quadrant, $\gamma $ lies in after one complete rotation again in first quadrant and $\delta $ lies in after one complete rotation again in second quadrant so first calculate the values of all the angles in terms of $\alpha $as above then substitute these values in the given equation and simplify after simplification use some basic trigonometric properties as above and again simplify as above, we will get the required answer.