Question

If A, B and C are interior angles of a triangle ABC, then show that sin²(A/2)+ sin² (B+C/2)=1.



${sin}^{2} ( \frac{a}{2} ) + {sin}^{2} ( \frac{b + c}{2} ) = 1$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that, A, B and C are interior angles of a triangle ABC.

So, using angle sum property, we get

$\rm \: A + B + C = 180 \degree$

$\rm \:B + C = 180 \degree - A$

$\rm \: \frac{B + C}{2} = \frac{180 \degree - A}{2}$

$\rm\implies \:\rm \: \frac{B + C}{2} = 90 \degree - \frac{A}{2}$

Now, Consider

$\rm \: {sin}^{2}\dfrac{A}{2} \: + \: {sin}^{2}\bigg(\dfrac{B + C}{2} \bigg)$

$\rm \: = \: {sin}^{2}\dfrac{A}{2} \: + \: {sin}^{2}\bigg(90 \degree - \frac{A}{2} \bigg)$

We know,

$\boxed{\rm{  \:sin(90 \degree - x) = cosx \: }}$

So, using this identity, we get

$\rm \: = \: {sin}^{2}\dfrac{A}{2} \: + \: {cos}^{2} \frac{A}{2}$

$\rm \: = \: 1$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: {sin}^{2}\dfrac{A}{2} \: + \: {sin}^{2}\bigg(\dfrac{B + C}{2} \bigg) = 1 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \\ \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$