Question

Within what respective limits must \[\dfrac{A}{2}\] lies when
1). \[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
2). \[2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
3). \[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \] and
4). \[2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \]

Answer 1

Hint: To solve this question we have to convert \[\sin A\] into the half-angle formula and put the values in the options and simplify all those. If the right-hand side is equal to the left-hand side then that option is the correct answer. Use the trigonometry property in the place of \[1\] then try to make it a perfect square and take them outside of the root.

Complete step-by-step solution:
Given,
Few options are given
\[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
\[2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
\[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \] and
\[2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \]
To find,
which option is correct;
Here, all these options take a look to the right hand side then we observe that only \[\sqrt {1 + \sin A} \] and \[\sqrt {1 - \sin A} \] are there. All other things are the different mathematical operators are used at different operators.
So, applying formula on only \[\sqrt {1 + \sin A} \] and \[\sqrt {1 - \sin A} \]
First we solve,
\[\sqrt {1 + \sin A} \]
Let this be \[a\]
\[a = \sqrt {1 + \sin A} \]
On putting the identity \[1 = {\sin ^2}\dfrac{A}{2} + {\cos ^2}\dfrac{A}{2}\] and half angle of \[\sin A\]
\[a = \sqrt {{{\sin }^2}\dfrac{A}{2} + {{\cos }^2}\dfrac{A}{2} + 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}} \] (\[\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\])
On making the terms in the form of \[{(a + b)^2}\]
\[a = \sqrt {{{(\sin \dfrac{A}{2} + \cos \dfrac{A}{2})}^2}} \]
Taking the terms outside of bracket
\[a = \sin \dfrac{A}{2} + \cos \dfrac{A}{2}\]
Again put the value of \[a\]
\[\sqrt {1 + \sin A} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2}\] …………………………(i)
Now we solve
\[\sqrt {1 - \sin A} \]
Let, this be \[b\]
\[b = \sqrt {1 - \sin A} \]
On putting the identity \[1 = {\sin ^2}\dfrac{A}{2} + {\cos ^2}\dfrac{A}{2}\] and half angle of \[\sin A\]
\[b = \sqrt {{{\sin }^2}\dfrac{A}{2} + {{\cos }^2}\dfrac{A}{2} - 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}} \] (\[\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\])
On making the terms in the form of \[{(a - b)^2}\]
\[b = \sqrt {{{(\sin \dfrac{A}{2} - \cos \dfrac{A}{2})}^2}} \]
Taking the terms outside of bracket
\[b = \sin \dfrac{A}{2} - \cos \dfrac{A}{2}\]
Again put the value of \[a\]
\[\sqrt {1 - \sin A} = \sin \dfrac{A}{2} - \cos \dfrac{A}{2}\] ……………………(ii)
Option 1.
\[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Putting the values form equation (i) and (ii)
\[2\sin \dfrac{A}{2} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2} + \sin \dfrac{A}{2} - \cos \dfrac{A}{2}\]
On further solving
\[2\sin \dfrac{A}{2} = 2\sin \dfrac{A}{2}\]
Option 1 is the correct answer
Option 2.
\[2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Putting the values form equation (i) and (ii)
\[2\sin \dfrac{A}{2} = - (\sin \dfrac{A}{2} + \cos \dfrac{A}{2}) + \sin \dfrac{A}{2} - \cos \dfrac{A}{2}\]
On further solving
\[2\sin \dfrac{A}{2} = - 2\cos \dfrac{A}{2}\]
Option 2 is not the correct answer
Option 3.
\[2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \]
Putting the values form equation (i) and (ii)
\[2\sin \dfrac{A}{2} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2} - (\sin \dfrac{A}{2} - \cos \dfrac{A}{2})\]
On further solving
\[2\sin \dfrac{A}{2} = 2\cos \dfrac{A}{2}\]
Option 3 is not the correct answer but the answer is matched with option 4 so option 4 is also the correct answer
Final answer:
Option 1 and option 4 both are correct.

Note: To solve these types of questions we have to use the property of trigonometry and must know all the identity and formulas of trigonometry and have a good practice to use all the formulas. In this particular question, we use two formulas of trigonometry. Make the term inside the root a perfect square.