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 positive.

Answer 1

Hint: We need to find out the required range for that first we need to solve the given formula, then apply the signs in different cases and then finally we will get the reduced form.
Comparing the formula \[2{\cos ^2}A - 1 = \cos 2A\] we will get the range for each case.

Formula used: \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
\[{(a - b)^2} = {a^2} - 2ab + {b^2}\]
\[{a^2} - {b^2} = (a + b)(a - b)\]
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[2{\cos ^2}A - 1 = \cos 2A\]

Complete step-by-step answer:
It is given that, the formula \[2\cos \dfrac{A}{2} = \pm \sqrt {1 + \sin A} \pm \sqrt {1 - \sin A} \],
We need to find out 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 positive.
Given that, \[2\cos \dfrac{A}{2} = \pm \sqrt {1 + \sin A} \pm \sqrt {1 - \sin A} \]
If both signs are taken positive, then
\[ \Rightarrow 2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Squaring both sides we have,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = {\left( {\sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
By using the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 1 + \sin A + 1 - \sin A + 2\sqrt {1 + \sin A} .\sqrt {1 - \sin A} \]
Simplifying we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)} \]
Let us using the formula \[{a^2} - {b^2} = (a + b)(a - b)\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {1 - {{\sin }^2}A} \]
Since, we know that $1 - {\sin ^2}x = {\cos ^{2x}}$ by using this we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {{{\cos }^2}A} \]
Cancelling common term \[2\] on both sides,
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1 + \left| {\cos A} \right|\]
We know that
\[2{\cos ^2}A - 1 = \cos 2A\]
Thus we get,
\[2{\cos ^2}\dfrac{A}{2} = 1 + \left| {\cos A} \right|\]
\[\left| {\cos A} \right| = 2{\cos ^2}\dfrac{A}{2} - 1\]
Then,\[\cos A > 0\] Thus \[cos{\text{ }}A\] is positive if A lie in the region \[ - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}\]
Therefore we get, \[\dfrac{A}{2}\] must lie between \[ - \dfrac{\pi }{4}{\text{ to }}\dfrac{\pi }{4}\] \[\]
 (2) If both signs are taken negative, then
\[2\cos \dfrac{A}{2} = - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} \]
Squaring both sides we have,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = {\left( { - \sqrt {1 + \sin A} - \sqrt {1 - \sin A} } \right)^2}\]
Taking common ${\left( { - 1} \right)^2}$ out from the root,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = {( - 1)^2}{\left( {\sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
By using the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 1 + \sin A + 1 - \sin A + 2\sqrt {1 + \sin A} .\sqrt {1 - \sin A} \]
Simplifying we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)} \]
Let us using the formula \[{a^2} - {b^2} = (a + b)(a - b)\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {1 - {{\sin }^2}A} \]
Since, we know that $1 - {\sin ^2}x = {\cos ^{2x}}$ by using this we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 + 2\sqrt {{{\cos }^2}A} \]
Cancelling common term \[2\] on both sides,
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1 + \left| {\cos A} \right|\]
We know that \[2{\cos ^2}A - 1 = \cos 2A\]
Thus we get,
\[2{\cos ^2}\dfrac{A}{2} = 1 + \left| {\cos A} \right|\]
\[\left| {\cos A} \right| = 2{\cos ^2}\dfrac{A}{2} - 1\]
Then, \[\cos A > 0\] Thus \[cos{\text{ }}A\] is positive if A lie in the region \[ - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}\]
Therefore we get, \[\dfrac{A}{2}\] must lie between \[ - \dfrac{\pi }{4}{\text{ to }}\dfrac{\pi }{4}\].
 (3) The first sign is negative and the second positive.
\[2\cos \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} \]
Squaring both sides we have,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = {\left( { - \sqrt {1 + \sin A} + \sqrt {1 - \sin A} } \right)^2}\]
By using the formula \[{(a - b)^2} = {a^2} - 2ab + {b^2}\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 1 + \sin A + 1 - \sin A - 2\sqrt {1 + \sin A} .\sqrt {1 - \sin A} \]
Simplifying we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2\sqrt {\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)} \]
Let us using the formula \[{a^2} - {b^2} = (a + b)(a - b)\],
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2\sqrt {1 - {{\sin }^2}A} \]
Since, we know that $1 - {\sin ^2}x = {\cos ^{2}x}$ by using this we get,
\[ \Rightarrow 4{\cos ^2}\dfrac{A}{2} = 2 - 2\sqrt {{{\cos }^2}A} \]
Cancelling common term \[2\] on both sides,
\[ \Rightarrow 2{\cos ^2}\dfrac{A}{2} = 1 - \left| {\cos A} \right|\]
\[ \Rightarrow \left| {\cos A} \right| = 1 - 2{\cos ^2}\dfrac{A}{2}\]
We know that \[2{\cos ^2}A - 1 = \cos 2A\]
If \[\cos A < 0\] then \[\left| {\cos A} \right| = - \cos A\]
 Thus,
 \[ \Rightarrow - \cos A = 1 - 2{\cos ^2}\dfrac{A}{2}\]
\[ \Rightarrow \cos A = 2{\cos ^2}\dfrac{A}{2} - 1\]
So, \[cos{\text{ }}A\] is negative if A lie in the region \[\dfrac{\pi }{2}{\text{ to }}\dfrac{{3\pi }}{2}\]
Therefore we get, \[\dfrac{A}{2}\] must lie between \[\dfrac{\pi }{4}{\text{ to }}\dfrac{{3\pi }}{4}\].

Note:
               

In the first quadrant \[\left( {0\,{\text{to }}\dfrac{\pi }{2}} \right)\] all trigonometric functions are positive, in second quadrant \[\left( {\dfrac{\pi }{2}\,{\text{to}}\,\pi } \right)\] only sine function is positive, and in third quadrant \[\left( {\pi \,{\text{to }}\dfrac{{3\pi }}{2}} \right)\] tan function is positive, in fourth quadrant
\[\left( {\dfrac{{3\pi }}{2}{\text{ to }}2\pi } \right)\] cosine functions are positive.