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Question: If \(\alpha ,\beta ,\gamma ,\delta \) are the smallest positive angles in ascending order of magnitu...

If α,β,γ,δ\alpha ,\beta ,\gamma ,\delta are the smallest positive angles in ascending order of magnitude which have their sines equal to positive quantity λ\lambda then the value of 4sin⁡α2+3sin⁡β2+2sin⁡γ2+sin⁡δ24\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2} is
A. 21−λ B. 21+λ C. 2λ D. 2λ+2  A.{\text{ }}2\sqrt {1 - \lambda } \\\ B.{\text{ }}2\sqrt {1 + \lambda } \\\ C.{\text{ }}2\sqrt \lambda \\\ D.{\text{ }}2\sqrt {\lambda + 2} \\\

Explanation

Solution

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 α<β<γ<δ\alpha < \beta < \gamma < \delta so, use this concept to reach the solution of the question.

Complete step-by-step answer:

Given condition is
α,β,γ,δ\alpha ,\beta ,\gamma ,\delta are the smallest positive angles in ascending order of magnitude which have their sines equal to positive quantity λ\lambda .
⇒sin⁡α=sin⁡β=sin⁡γ=sin⁡δ=λ\Rightarrow \sin \alpha = \sin \beta = \sin \gamma = \sin \delta = \lambda………………. (1)
Then we have to find out the value of
4sin⁡α2+3sin⁡β2+2sin⁡γ2+sin⁡δ24\sin \dfrac{\alpha }{2} + 3\sin \dfrac{\beta }{2} + 2\sin \dfrac{\gamma }{2} + \sin \dfrac{\delta }{2}…………………. (2)
Now from equation (1)
sin⁡α=sin⁡β\sin \alpha = \sin \beta
⇒sin⁡β=sin⁡(π−α)\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 α<β<γ<δ\alpha < \beta < \gamma < \delta ).
⇒β=π−α\Rightarrow \beta = \pi - \alpha
Similarly (γ=2π+α),(δ=3π−α)\left( {\gamma = 2\pi + \alpha } \right),\left( {\delta = 3\pi - \alpha } \right).
Now substitute all these values in equation (2) we have
⇒4sin⁡α2+3sin⁡π−α2+2sin⁡2π+α2+sin⁡3π−α2\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
⇒4sin⁡α2+3sin⁡(π2−α2)+2sin⁡(π+α2)+sin⁡(3π2−α2)\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 [sin⁡(π2−θ)=cos⁡θ,sin⁡(π+θ)=−sin⁡θ,sin⁡(3π2−θ)=−cos⁡θ]\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,
⇒4sin⁡α2+3cos⁡α2−2sin⁡α2−cos⁡α2\Rightarrow 4\sin \dfrac{\alpha }{2} + 3\cos \dfrac{\alpha }{2} - 2\sin \dfrac{\alpha }{2} - \cos \dfrac{\alpha }{2}
⇒2(sin⁡α2+cos⁡α2)\Rightarrow 2\left( {\sin \dfrac{\alpha }{2} + \cos \dfrac{\alpha }{2}} \right)…………………………. (3)
Now as we know (sin⁡α2+cos⁡α2)2=sin⁡2α2+cos⁡2α2+2sin⁡α2cos⁡α2{\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
⇒(sin⁡α2+cos⁡α2)2=1+sin⁡α ⇒(sin⁡α2+cos⁡α2)=1+sin⁡α  \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,

⇒4sin⁡α2+3sin⁡β2+2sin⁡γ2+sin⁡δ2=2(sin⁡α2+cos⁡α2)=21+sin⁡α\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⁡α=λ\sin \alpha = \lambda
⇒4sin⁡α2+3sin⁡β2+2sin⁡γ2+sin⁡δ2=21+λ\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 α<β<γ<δ\alpha < \beta < \gamma < \delta and it is also given that sin⁡α=sin⁡β=sin⁡γ=sin⁡δ=λ\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = \lambda so, according to property of sine which is stated above, α\alpha lies in first quadrant, β\beta lies in second quadrant, γ\gamma lies in after one complete rotation again in first quadrant and δ\delta lies in after one complete rotation again in second quadrant so first calculate the values of all the angles in terms of α\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.