A) \[2\sqrt{1-k}\]
B) \[2\sqrt{1-k}\]
C) \[\frac{\sqrt{1+k}}{2}\]
D) \[\sqrt{1+k}\]
Correct Answer: B
Solution :
Given, \[\alpha <\beta <\gamma <\delta \] Also, \[\sin \,\alpha =\sin \beta =\sin \gamma =\sin \delta =k\] \[\therefore \] \[\beta =\pi -\alpha ,\,\,\gamma =2\pi +\alpha ,\,\,\delta =3\pi -\alpha \] Now, \[4\,\,\sin \frac{\alpha }{2}+3\,\,\sin \frac{\beta }{2}+2\,\sin \frac{\gamma }{2}+\sin \frac{\delta }{2}\] \[=4\sin \frac{\alpha }{2}+3\sin \left( \frac{\pi -\alpha }{2} \right)+2\sin \left( \frac{2\pi +\alpha }{2} \right)\] \[+\sin \left( \frac{3\pi -\alpha }{2} \right)\] \[=4\sin \frac{\alpha }{2}+3\,\cos \frac{\alpha }{2}-2\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2}\] \[=2\sin \frac{\alpha }{2}+2\cos \frac{\alpha }{2}\] \[=2\sqrt{{{\left( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} \right)}^{2}}}\] \[=2\sqrt{{{\sin }^{2}}\frac{\alpha }{2}+{{\cos }^{2}}\frac{\alpha }{2}+2\,\sin \frac{\alpha }{2}\,\cos \frac{\alpha }{2}}\] \[=2\sqrt{1+\sin \,\alpha }\] \[=2\sqrt{1+k}\]
You need to login to perform this action.
You will be redirected in
3 sec
