Multiples And Sub-Multiples
Theorem :
Prove that i)\(
\sin 22\frac{1}
{2}^0 = \sqrt {\frac{{\sqrt 2 - 1}}
{{2\sqrt 2 }}}
\) ii)\(
\cos 22\frac{1}
{2}^0 = \sqrt {\frac{{\sqrt 2 + 1}}
{{2\sqrt 2 }}}
\) iii)\(
\tan 22\frac{1}
{2}^0 = \sqrt 2 - 1
\)
Proof : We know,
\(
\sin \frac{A}
{2} = \pm \sqrt {\frac{{1 - \cos A}}
{2}}
\)
put A = 450
\(
\sin 22\frac{1}
{2}^0 = \sqrt {\frac{{1 - \cos 45^0 }}
{2}}
\)
=\(
\sqrt {\frac{{1 - \frac{1}
{{\sqrt 2 }}}}
{2}}
\)
\(
\sin 22\frac{1}
{2}^0 = \sqrt {\frac{{\sqrt 2 - 1}}
{{2\sqrt 2 }}}
\)
ii) \(
\cos \frac{A}
{2} = \sqrt {\frac{{1 + \cos A}}
{2}}
\)
put A =450
\(
\cos 22\frac{1}
{2}^0 = \sqrt {\frac{{1 + \cos 45^0 }}
{2}}
\)
=\(
\sqrt {\frac{{1 + \frac{1}
{{\sqrt 2 }}}}
{2}}
\)
\(
\sin 22\frac{1}
{2}^0 = \sqrt {\frac{{\sqrt 2 - 1}}
{{2\sqrt 2 }}}
\)
\(
\cos 22\frac{1}
{2}^0 = \sqrt {\frac{{\sqrt 2 + 1}}
{{2\sqrt 2 }}}
\)
iii) \(
\tan 22\frac{1}
{2}^0 = \sqrt {\frac{{1 - \cos 2A}}
{{1 + \cos 2A}}}
\)
=\(
\sqrt {\frac{{1 - \frac{1}
{{\sqrt 2 }}}}
{{1 + \frac{1}
{{\sqrt 2 }}}}}
\) =\(
\sqrt {\frac{{\sqrt 2 - 1}}
{{\sqrt 2 + 1}}}
\)
\(
\sqrt {\frac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}
{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}}}
\)\(
\sqrt {\frac{{\left( {\sqrt 2 - 1} \right)^2 }}
{{\sqrt 2 ^2 - 1^2 }}}
\)
\(
\tan 22\frac{1}
{2}^0 = \sqrt 2 - 1
\)
Multiples And Sub-Multiples
Theorem
Prove that:i)\(
\sin 7\frac{1}
{2}^0 = \frac{{\sqrt {4 - \sqrt 6 - \sqrt 2 } }}
{{2\sqrt 2 }}
\) ii)\(
\cos 7\frac{1}
{2}^0 = \frac{{\sqrt {4 + \sqrt 6 + \sqrt 2 } }}
{{2\sqrt 2 }}
\)
iii)\(
\tan 7\frac{1}
{2}^0 = \sqrt 2 - \sqrt 3 - \sqrt 4 + \sqrt 6 = \left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 2 - 1} \right)
\)
iv)\(
\cot 7\frac{1}
{2}^0 = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = \left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 2 + 1} \right)
\)
Proof : We have
\(
\cos 15^0 = \frac{{\sqrt 3 + 1}}
{{2\sqrt 2 }}
\)
we know
\(
\sin \frac{A}
{2} = \pm \sqrt {\frac{{1 - \cos A}}
{2}}
\)
\(
= \, \pm \sqrt {\frac{{1 - \left( {\frac{{\sqrt 3 + 1}}
{{2\sqrt 2 }}} \right)}}
{2}}
\)
\(
= \, \pm \sqrt {\frac{{2\sqrt 2 - \sqrt 3 - 1}}
{{4\sqrt 2 }}}
\) (putting A=150)
\(
\sin 7\frac{1}
{2}^0 = \frac{{\sqrt {4 - \sqrt 6 - \sqrt 2 } }}
{{2\sqrt 2 }}
\) (\(
\because
\) \(
\sin 7\frac{1}
{2}\,\,is + ve
\))…….(1)
ii)We have,
\(
\cos \frac{A}
{2} = \pm \sqrt {\frac{{1 + \cos A}}
{2}}
\)
putting A =150
\(
\cos 7\frac{1}
{2}\,\,\, = \,\,\, \pm \,\,\,\,\sqrt {\frac{{1 + \left( {\frac{{\sqrt 3 + 1}}
{{2\sqrt 2 }}} \right)}}
{2}}
\)
\(
\cos 7\frac{1}
{2} = \sqrt {\frac{{2\sqrt 2 + \sqrt 3 + 1}}
{{4\sqrt 2 }}}
\)
\(
= \sqrt {\frac{{4 + \sqrt 6 + \sqrt 2 }}
{8}}
\)
\(
\cos 7\frac{1}
{2}^0 = \frac{{\sqrt {4 + \sqrt 6 + 2\sqrt 2 } }}
{{2\sqrt 2 }}
\)……….(2)
iii)\(
\tan 7\frac{1}
{2}^0 = \frac{{\sin 7\frac{1}
{2}^0 }}
{{\cos 7\frac{1}
{2}^0 }}
\)
=\(
\frac{{2\sin ^2 7\frac{1}
{2}^0 }}
{{2\sin 7\frac{1}
{2}^0 \,.\,\cos 7\frac{1}
{2}^0 }}
\)=\(
\frac{{1 - \cos 15^0 }}
{{\sin 15^0 }}
\) \(
\left( \begin{gathered}
1 - \cos 2\theta = 2\sin ^2 \theta \hfill \\
\sin 2\theta = 2\sin \theta \,\,\,\,cos\theta \hfill \\
\end{gathered} \right)
\)
=\(
\frac{{1 - \left( {\frac{{\sqrt 3 + 1}}
{{2\sqrt 2 }}} \right)}}
{{\left( {\frac{{\sqrt 3 - 1}}
{{2\sqrt 2 }}} \right)}}
\)
=\(
\frac{{2\sqrt 2 - \sqrt 3 - 1}}
{{\sqrt 3 - 1}}
\)
=\(
\frac{{\left( {2\sqrt 2 - \sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}
{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}
\)
=\(
\frac{{2\sqrt 6 + 2\sqrt 2 - 3 - \sqrt 3 - \sqrt 3 - 1}}
{2}
\)
=\(
\frac{{2\sqrt 6 - 2\sqrt 3 + 2\sqrt 2 - 4}}
{2}
\)
=\(
\frac{{2\left( {\sqrt 6 - \sqrt 3 + \sqrt 2 - 2} \right)}}
{2}
\)
=\(
\sqrt 6 - \sqrt 3 + \sqrt 2 - 2
\)
\(
\tan 7\frac{1}
{2}^0 = \left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 2 - 1} \right)
\)
iv)\(
\cot 7\frac{1}
{2}^0 = \frac{{\cos 7\frac{1}
{2}^0 }}
{{\sin 7\frac{1}
{2}^0 }}
\)
=\(
\frac{{2\cos ^2 7\frac{1}
{2}^0 }}
{{2\sin 7\frac{1}
{2}^0 .\,\cos 7\frac{1}
{2}^0 }}
\)
=\(
\frac{{1 + \cos 15^0 }}
{{\sin 15^0 }}
\)=\(
\frac{{1 + \left( {\frac{{\sqrt 3 + 1}}
{{2\sqrt 2 }}} \right)}}
{{\left( {\frac{{\sqrt 3 - 1}}
{{2\sqrt 2 }}} \right)}}
\)=