Question

if sin theta = a/b+c then show that cos theta =2√bc/b+c cosA/2​

Answer 1

Step-by-step explanation:

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Answer 2

Given,

$\[\sin \theta = \frac{a}{{b + c}}\]$

Show that,

$\[\cos \theta = \frac{{2\sqrt {bc} }}{{b + c}}\cos \frac{A}{2}\]$

Solution,

$\[\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \]$

$\[ \Rightarrow \cos \theta = \sqrt {1 - {{\left( {\frac{a}{{b + c}}} \right)}^2}} \]$

$\[ \Rightarrow \cos \theta = \sqrt {\frac{{{b^2} + {c^2} + 2bc - {a^2}}}{{{{\left( {b + c} \right)}^2}}}} \]$

$\[ \Rightarrow \cos \theta = \frac{{\sqrt {{b^2} + {c^2} - {a^2} + 2bc} }}{{b + c}}\]$

Know that, $\[\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]$

$\[ \Rightarrow \cos \theta = \frac{{\sqrt {2bc\cos A + 2bc} }}{{b + c}}\]$

$\[ \Rightarrow \cos \theta = \frac{{\sqrt {2bc\left( {1 + \cos A} \right)} }}{{b + c}}\]$

$\[ \Rightarrow \cos \theta = \frac{{\sqrt {2bc\left( {1 + 2{{\cos }^2}\left( {\frac{A}{2}} \right) - 1} \right)} }}{{b + c}}\]$

$\[ \Rightarrow \cos \theta = \frac{{\sqrt {2bc \times \left( {2{{\cos }^2}\left( {\frac{A}{2}} \right)} \right)} }}{{b + c}}\]$

$\[ \Rightarrow \cos \theta = \frac{{2\sqrt {bc} }}{{b + c}}\cos \left( {\frac{A}{2}} \right)\]$

Hence the given condition is proved.