Question

In the formula \[2\cos \dfrac{A}{2} = \pm \sqrt {1 + \sin A} \pm \sqrt {1 - \sin A} \], find within what limits $\dfrac{A}{2}$ must lie when
1) The two positive signs are taken,
2) The two negative signs are taken, and
3) The first sign is negative and the second is positive.

Answer 1

Hint:We will first take as given in each part and then square both the sides of the resultant expressions. Then, using some trigonometric identities, we will simplify them and see when LHS will be equal to RHS and thus, we will have our answer.

Complete step-by-step answer:
Part 1) when both signs are taken to be positive.
The expression becomes: \[2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Squaring both sides of the expression, we will get:-
\[ \Rightarrow {\left( {2\cos \dfrac{A}{2}} \right)^2} = {\left( {\sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
Now, to simplify RHS, we will use the formula: ${(a + b)^2} = {a^2} + {b^2} + 2ab$.
So, we will get:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 1 + \sin A + 1 - \sin A + 2\sqrt {(1 + \sin A)(1 - \sin A)} \]
Now, we will use the formula: $(a + b)(a - b) = {a^2} - {b^2}$.
We will get:- \[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {1 - {{\sin }^2}A} \]
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Hence, our expression becomes:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {{{\cos }^2}A} \]
We also know that: $\sqrt {{x^2}} = |x|$.
Hence, we get:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2|\cos A|\]
Taking out 2 and cutting it off from both sides:-
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1 + |\cos A|\]
Rearranging the terms to get:-
\[ \Rightarrow |\cos A| = 2{\cos ^2}\dfrac{A}{2} - 1\] ………..(A)
Now, we also have the identity: $2{\cos ^2}\theta - 1 = \cos 2\theta $
We see that the RHS in (A) is tending to be $\cos A$.
Hence, the LHS needs to be positive as well.
That means we need $\cos A > 0$.
Since, the principal value of ${\cos ^{ - 1}}x$ lies in $[0,\pi ]$ among which the positive part takes the value: $\left[ {0,\dfrac{\pi }{2}} \right]$.
Therefore, $A \in \left[ {0,\dfrac{\pi }{2}} \right]$ and thus $\dfrac{A}{2} \in \left[ {0,\dfrac{\pi }{4}} \right]$.

Part 2) when both signs are taken to be negative.
The expression becomes: \[2\cos \dfrac{A}{2} = - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \]
Squaring both sides of the expression, we will get:-
\[ \Rightarrow {\left( {2\cos \dfrac{A}{2}} \right)^2} = {\left( { - 1} \right)^2}{\left( {\sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
On rewriting it by simplifying a bit, we will get:-
\[ \Rightarrow {\left( {2\cos \dfrac{A}{2}} \right)^2} = {\left( {\sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
Now, this is the same as part 1), hence, we will get the same answer.
Therefore, $\dfrac{A}{2} \in \left[ {0,\dfrac{\pi }{4}} \right]$.

Part 3) when the first sign is negative and second is positive.
The expression becomes: \[2\cos \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Squaring both sides of the expression, we will get:-
\[ \Rightarrow {\left( {2\cos \dfrac{A}{2}} \right)^2} = {\left( { - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
Now, to simplify RHS, we will use the formula: ${(a - b)^2} = {a^2} + {b^2} - 2ab$.
So, we will get:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 1 + \sin A + 1 - \sin A - 2\sqrt {(1 + \sin A)(1 - \sin A)} \]
Now, we will use the formula: $(a + b)(a - b) = {a^2} - {b^2}$.
We will get:- \[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2\sqrt {1 - {{\sin }^2}A} \]
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Hence, our expression becomes:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2\sqrt {{{\cos }^2}A} \]
We also know that: $\sqrt {{x^2}} = |x|$.
Hence, we get:
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2|\cos A|\]
Taking out 2 and cutting it off from both sides:-
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1 - |\cos A|\]
Rearranging the terms to get:-
\[ \Rightarrow |\cos A| = 1 - 2{\cos ^2}\dfrac{A}{2}\] ………..(B)
Now, we also have the identity: $2{\cos ^2}\theta - 1 = \cos 2\theta $
We see that the RHS in (B) is tending to be $ - \cos A$.
Hence, the LHS needs to be negative as well.
That means we need $\cos A < 0$.
Since, the principal value of ${\cos ^{ - 1}}x$ lies in $[0,\pi ]$ among which the negative part takes the value: $\left[ {\dfrac{\pi }{2},\pi } \right]$.
Therefore, $A \in \left[ {\dfrac{\pi }{2},\pi } \right]$ and thus $\dfrac{A}{2} \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]$.

Note:The students must note that when we cut off 2 by taking it common on both sides, we could do that because 2 is not equal to 0. We can never cut a variable from both sides which may possess the value 0.
The second part was similar as first part because:
\[{( - a - b)^2} = {[( - a) + ( - b)]^2} = {( - a)^2} + {( - b)^2} + 2( - a)( - b) = {a^2} + {b^2} + 2ab = {(a + b)^2}\]