Question

If $2tan\frac{\alpha }{2}=tan\frac{\beta }{2},$ prove that $cos\alpha =\frac{3+5cos\beta }{5+3cos\beta }.$ A.so, determine the value of $cos\alpha$ when $\beta ={60}^{\circ }.$

Or

If a $sin\theta =bsin(\theta +\frac{2\pi }{3})=csin(\theta +\frac{4\pi }{3}),$ find the value of ab + bc + ca.

Answer 1

We have, $RHS=\frac{3+5cos\beta }{5+3cos\beta }$

= $\frac{3+5\left(\frac{1-ta{n}^2\beta /2}{1+ta{n}^2\beta /2}\right)}{5+3\left(\frac{1-ta{n}^2\beta /2}{1+ta{n}^2\beta /2}\right)}[\because cos\beta =\frac{1-ta{n}^2\beta /2}{1+ta{n}^2\beta /2}]$

= $\frac{3+3ta{n}^2\beta /2+5-5ta{n}^2\beta /2}{5+5ta{n}^2\beta /2+3-3ta{n}^2\beta /2}$

= $\frac{8-2ta{n}^2\beta /2}{8+2ta{n}^2\beta /2}=\frac{8-2\left(4ta{n}^2\alpha /2\right)}{8+2\left(4ta{n}^2\alpha /2\right)}$

$[\because 2tan\alpha /2=tan\beta /2]$

= $\frac{8-8ta{n}^2\alpha /2}{8+8ta{n}^2\alpha /2}$

= $\frac{1-ta{n}^2\alpha /2}{1+ta{n}^2\alpha /2}=cos\alpha =LHS$

Hence proved.

Now, when $\beta ={60}^{\circ },$ then

$cos\alpha =\frac{3+5cos{60}^{\circ }}{5+3cos{60}^{\circ }}=\frac{3+5\times 1/2}{5+3\times 1/2}$

= $\frac{6+5}{10+3}=\frac{11}{13}$

Or

We have, $asin\theta =bsin(\theta +\frac{2\pi }{3})=csin(\theta +\frac{4\pi }{3})=k$ (say)

Then, $sin\theta =\frac{k}{a},sin(\theta +\frac{2\pi }{3})=\frac{k}{b}$

and $sin(\theta +\frac{4\pi }{3})=\frac{k}{c}$

Now, $ab+bc+ca=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

= $\frac{abc}{k}[sin\theta +sin(\theta +\frac{2\pi }{3})+sin(\theta +\frac{4\pi }{3})]$

= $\frac{abc}{k}[sin\theta +sin\theta cos\frac{2\pi }{3}+cos\theta sin\frac{2\pi }{3}+sin\theta cos\frac{4\pi }{3}+cos\theta sin\frac{4\pi }{3}]$

= $[\because sin(A+B)=sinAcosB+cosAsinB]$

= $\frac{abc}{k}(sin\theta -\frac{1}{2}sin\theta +\frac{\sqrt{3}}{2}cos\theta -\frac{1}{2}sin\theta -\frac{\sqrt{3}}{2}cos\theta )$

$[\because cos\frac{2\pi }{3}=cos(\pi -\frac{\pi }{3})=-cos\frac{\pi }{3}=-\frac{1}{2^{\prime }}sin\frac{2\pi }{3}=sin(\pi -\frac{\pi }{3})=sin\frac{\pi }{3}=\frac{\sqrt{3}}{2},cos\frac{4\pi }{3}=cos(\pi +\frac{\pi }{3})=-cos\frac{\pi }{3}=-\frac{1}{2^{\prime }}sin\frac{4\pi }{3}=sin(\pi +\frac{\pi }{3})=-sin\frac{\pi }{3}=-\frac{\sqrt{3}}{2}]$

= $\frac{abc}{k}.0=0$

Hence, the value of ab + bc + ca is 0.