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May Lord ... [Backstage]

You must be familiar with the formulas:

\(\sin { A/2 } +\cos { A/2 } =\pm \sqrt { 1+\sin { A } }\)

\(\sin { A/2 } -\cos { A/2 } =\pm \sqrt { 1-\sin { A } } \)

The Problem May Lord... is related to problems such as expressing trigonometrical ratios of angle A/2 in terms of \(\sin { A }\).

There is always an ambiguity in such cases.

What I am trying to say is that knowing the value of \(\sin { A }\) does not uniquely determine the value of \(\sin { A/2 }\) and \(\cos { A/2 }\) but only gives the magnitude or absolute value of \(\sin { A/2 }\) and \(\cos { A/2 }\).To obtain the trigonometrical ratios completely,of A/2,in terms of \(\sin { A }\) we also need to find its sign [ +,-]. But,to determine the sign [ +,-] we need to know the quadrant in which the angle lies.

To find the ambiguities we can proceed d as follows:

\(\sin { A/2 } +\cos { A/2 } \)

\(=\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } \sin { A/2 } +\frac { 1 }{ \sqrt { 2 } } \cos { A/2 } \right) \)

\(=\sqrt { 2 } \sin { (\pi /4+A/2 } )\quad \) ...(i)

(i) is positive if\( A/2+\pi /4\) lies between \(2n\pi\) and \(2n\pi +\pi\)

i.e A/2 lies between \(2n\pi -\pi /4\) and \(2n\pi +3\pi /4\)

\(\therefore\) \(\sin { A/2 } +\cos { A/2 }\) is positive if

A/2 lies between \(2n\pi -\pi /4\) and \(2n\pi +3\pi /4\)

It is negative otherwise. [i.e between \(2n\pi +3\pi /4\) and \(2n\pi+7\pi /4\) ]

In just similar way it can be shown that \(sin { A/2 } -cos { A/2 } \) is positive

if A/2 lies between \(2n\pi +\pi /4\) and \(2n\pi +5\pi /4\).

It is negative otherwise. [i.e between \(2n\pi -3\pi /4\) and \(2n\pi +\pi /4\) ]

It will be much better to understand if you have a diagram[I have a diagram but I don't know how to upload it here,sorry for that].Draw the renowned four quadrants then mark of \(\pi /4,3\pi /4,5\pi /4,7\pi /4\).

This is all you require to shoot May Lord forgive us our sins

Some Examples:

  • Q# In the formula \(2\cos { A/2 } =\pm \sqrt { 1+\sin { A } } \pm \sqrt { 1-\sin { A } } \)

Find within what limits A\2 must lie when:

1.the two positive signs are taken

2.the two negative signs are taken

solution: 1. The formula \(2\cos { A/2 } =+\sqrt { 1+\sin { A } } +\sqrt { 1-\sin { A } } \)

Is obtained using the following two formulas:

\(\sin { A/2 } +\cos { A/2 } =+\sqrt { 1+\sin { A } } \\ \sin { A/2 } -\cos { A/2 } =-\sqrt { 1-\sin { A } } \)

Now \( \sin { A/2 } +\cos { A/2 }\) is +ve between \(2n\pi -\pi /4\) and \(2n\pi +3\pi /4\) and \( \sin { A/2 } -\cos { A/2 }\) is -ve between \(2n\pi -3\pi /4\) and \( 2n\pi +\pi /4\).Take the intersection.

Hence A/2 must lie within\(2n\pi -\pi /4\) and \( 2n\pi +\pi /4\).

2.The formula \(2\cos { A/2 } =-\sqrt { 1+\sin { A } } -\sqrt { 1-\sin { A } } \)

Is obtained using the following two formulas:

\(sin { A/2 } +\cos { A/2 } =-\sqrt { 1+\sin { A } } \\ \sin { A/2 } -\cos { A/2 } =+\sqrt { 1-\sin { A } } \)

Now \(\sin { A/2 } +\cos { A/2 }\) is -ve between \(2n\pi+3\pi /4\) and \(2n\pi+7\pi /4\) and \(\sin { A/2 } -\cos { A/2 }\) is +ve between \(2n\pi+\pi /4\) and \(2n\pi+5\pi /4\). Take the intersection.

Hence A/2 must lie within \( 2n\pi+3\pi /4\) and \(2n\pi+5\pi /4\).

Note by Soumo Mukherjee
2 years, 5 months ago

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