May Lord ... [Backstage]

You must be familiar with the formulas:

sinA/2+cosA/2=±1+sinA\sin { A/2 } +\cos { A/2 } =\pm \sqrt { 1+\sin { A } }

sinA/2cosA/2=±1sinA\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 sinA\sin { A }.

There is always an ambiguity in such cases.

What I am trying to say is that knowing the value of sinA\sin { A } does not uniquely determine the value of sinA/2\sin { A/2 } and cosA/2\cos { A/2 } but only gives the magnitude or absolute value of sinA/2\sin { A/2 } and cosA/2\cos { A/2 }.To obtain the trigonometrical ratios completely,of A/2,in terms of sinA\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:

sinA/2+cosA/2\sin { A/2 } +\cos { A/2 }

=2(12sinA/2+12cosA/2)=\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } \sin { A/2 } +\frac { 1 }{ \sqrt { 2 } } \cos { A/2 } \right)

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

(i) is positive ifA/2+π/4 A/2+\pi /4 lies between 2nπ2n\pi and 2nπ+π2n\pi +\pi

i.e A/2 lies between 2nππ/42n\pi -\pi /4 and 2nπ+3π/42n\pi +3\pi /4

\therefore sinA/2+cosA/2\sin { A/2 } +\cos { A/2 } is positive if

A/2 lies between 2nππ/42n\pi -\pi /4 and 2nπ+3π/42n\pi +3\pi /4

It is negative otherwise. [i.e between 2nπ+3π/42n\pi +3\pi /4 and 2nπ+7π/42n\pi+7\pi /4 ]

In just similar way it can be shown that sinA/2cosA/2sin { A/2 } -cos { A/2 } is positive

if A/2 lies between 2nπ+π/42n\pi +\pi /4 and 2nπ+5π/42n\pi +5\pi /4.

It is negative otherwise. [i.e between 2nπ3π/42n\pi -3\pi /4 and 2nπ+π/42n\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 π/4,3π/4,5π/4,7π/4\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 2cosA/2=±1+sinA±1sinA2\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 2cosA/2=+1+sinA+1sinA2\cos { A/2 } =+\sqrt { 1+\sin { A } } +\sqrt { 1-\sin { A } }

Is obtained using the following two formulas:

sinA/2+cosA/2=+1+sinAsinA/2cosA/2=1sinA\sin { A/2 } +\cos { A/2 } =+\sqrt { 1+\sin { A } } \\ \sin { A/2 } -\cos { A/2 } =-\sqrt { 1-\sin { A } }

Now sinA/2+cosA/2 \sin { A/2 } +\cos { A/2 } is +ve between 2nππ/42n\pi -\pi /4 and 2nπ+3π/42n\pi +3\pi /4 and sinA/2cosA/2 \sin { A/2 } -\cos { A/2 } is -ve between 2nπ3π/42n\pi -3\pi /4 and 2nπ+π/4 2n\pi +\pi /4.Take the intersection.

Hence A/2 must lie within2nππ/42n\pi -\pi /4 and 2nπ+π/4 2n\pi +\pi /4.

2.The formula 2cosA/2=1+sinA1sinA2\cos { A/2 } =-\sqrt { 1+\sin { A } } -\sqrt { 1-\sin { A } }

Is obtained using the following two formulas:

sinA/2+cosA/2=1+sinAsinA/2cosA/2=+1sinAsin { A/2 } +\cos { A/2 } =-\sqrt { 1+\sin { A } } \\ \sin { A/2 } -\cos { A/2 } =+\sqrt { 1-\sin { A } }

Now sinA/2+cosA/2\sin { A/2 } +\cos { A/2 } is -ve between 2nπ+3π/42n\pi+3\pi /4 and 2nπ+7π/42n\pi+7\pi /4 and sinA/2cosA/2\sin { A/2 } -\cos { A/2 } is +ve between 2nπ+π/42n\pi+\pi /4 and 2nπ+5π/42n\pi+5\pi /4. Take the intersection.

Hence A/2 must lie within 2nπ+3π/4 2n\pi+3\pi /4 and 2nπ+5π/42n\pi+5\pi /4.

Note by Soumo Mukherjee
5 years ago

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