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 \(2n\pi+5\pi /4\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

There are no comments in this discussion.