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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, 3 months ago