Question

In $△ABC$, if $B=3C$, prove that $cosC=\sqrt{\left(\frac{b+c}{2c}\right)}$, $sinC=\sqrt{\left(\frac{3c-b}{4c}\right)}$ and $sin\frac{A}{2}=\frac{b-c}{2c}$

Answer 1

$\frac{b+c}{4c}=\frac{sinB+sinC}{4sinC}$

$=\frac{sin3C+sinC}{4sinC}=\frac{2sin2CcosC}{4sinC}$

$=\frac{4sinCcosCcosC}{4sinC}=co{s}^{2}C.$

$\therefore cosC=\sqrt{(b+c)/4c}$

$\therefore sinC=\sqrt{(1-co{s}^{2}C)}$ etc.

Again $\frac{b-c}{2c}=\frac{sinB-sinC}{2sinC}$

$=\frac{sin3C-sinC}{2sinC}=\frac{2cos2CsinC}{2sinC}=cos2C$.

But $A+B+C={180}^0$ and $B=3C$

$\Rightarrow A+4C={180}^0$ and $B=3C$

$\Rightarrow A+4C={180}^0\Rightarrow 4C={180}^0-A$

or $2C={90}^0-\frac{1}{2}A.$

Hence $\frac{b-c}{2c}=cos({90}^0-\frac{A}{2})=sin\frac{A}{2}$

Alt. $\frac{sinb}{sinc}=\frac{b}{c}$ or $\frac{sin3C}{sinC}=\frac{b}{c}$

or $3-4si{n}^{2}C\frac{b}{c}\therefore sinC=\sqrt{\frac{3c-b}{4c}}$

Again $A={180}^0-(B+C)={180}^0-4C$.

$\therefore A/2={90}^0-2C$,

$\therefore sin\left(A/2\right)={90}^0-2C$,

$\therefore sin\left(A/2\right)=cos2C=1-si{n}^{2}C$ etc.