Question

Determine the value of (I) $\sin 67{\frac{1}{2}}^{\circ }$ (ii) $\cos 67\frac{1^{\circ }}{2}$ and (iii) $tan82\frac{1^{\circ }}{2}$

Answer 1

(i) $\sin 67\frac{1}{2}=\sin (90-22\frac{1}{2})=\cos 22\frac{1}{2}$

$\Rightarrow \sqrt{\frac{1+\cos 2A}{2}}=\sqrt{\frac{1+\cos 45}{2}}=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}$

$\Rightarrow \sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

(ii)$\cos 67\frac{1}{2}=\cos (90-22\frac{1}{2})=\sin 22\frac{1}{2}$

$\Rightarrow \sqrt{\frac{1-\cos 2A}{2}}$=$\sqrt{\frac{1-\cos 45}{2}}$=$\sqrt{\frac{1-\frac{1}{2}}{2}}$

$\Rightarrow \sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$

(iii) $tan82\frac{1}{2}=\tan \left(\frac{165}{2}\right)=\tan (180-15)=-\tan 15$

$=-\tan (45-30)$

$=-\left[\frac{\tan 45-\tan 30}{1+\tan 45\left(\tan 30\right)}\right]$

$=-\left[\frac{1-\frac{1}{\sqrt{3}}}{1+1\frac{1}{\sqrt{3}}}\right]$

$\Rightarrow -\left[\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\right]=\frac{1-\sqrt{3}}{1+\sqrt{3}}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}$

$=\frac{\sqrt{3}-(\sqrt{3}{)}^2-1+\sqrt{3}}{(\sqrt{3}{)}^2-1^2}$

$=\frac{2\sqrt{3}-3-1}{3-1}=\frac{2\sqrt{3}-4}{2}$

$\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\Rightarrow 2\theta =165$

$\Rightarrow \theta =\frac{165}{2}$

$\tan \left(165\right)=\sqrt{3}-2$

$\Rightarrow \tan \left(165\right)=\frac{2\tan \left(\frac{165}{2}\right)}{1-{\tan }^2\left(\frac{165}{2}\right)}$

Let $\tan \frac{165}{2}=x$

$\Rightarrow \sqrt{3}-2=\frac{2x}{1-x^2}$

$\Rightarrow (1-x^2)(\sqrt{3}-2)=2x$

$\Rightarrow (\sqrt{3}-2)-x^2(\sqrt{3}-2)=2x$

$\Rightarrow (2-\sqrt{3})x^2-2x+(\sqrt{3}-2)=0$

$\Rightarrow x=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(2-\sqrt{3}\left)\right(\sqrt{3}-2)}}{2(2-\sqrt{3})}$

$x=\frac{2\pm \sqrt{4+4(2-\sqrt{3})(2-\sqrt{3})}}{2(2-\sqrt{3})}$

$=\frac{2\pm 2\sqrt{1+(2{)}^2+(\sqrt{3}{)}^2-2\times 2\sqrt{3}}}{(2-\sqrt{3})}$

$=\frac{1\pm \sqrt{1+4+3-4\sqrt{3}}}{2-\sqrt{3}}$

$\because tan84\frac{1}{2}\angle 90$

$\Rightarrow \tan 84\frac{1}{2}$

$=+ve{1}^{st}quad$

$x=\frac{1+\sqrt{8-4\sqrt{3}}}{2-\sqrt{3}}$

$=\frac{1+2\sqrt{2-\sqrt{3}}}{2-\sqrt{3}}\times \frac{(2+\sqrt{3})}{2+\sqrt{3}}$

$=\frac{(2+\sqrt{3})+2\sqrt{2-\sqrt{3}}(2+\sqrt{3})}{4-3}$

$=2+\sqrt{3}+2(2+\sqrt{3})\sqrt{2-\sqrt{3}}$

$=2+\sqrt{3}+\left[2\right(2+\sqrt{3}\left)\right(\sqrt{2-\sqrt{3}}\left)\right]\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$

$=2+\sqrt{3}+2\sqrt{2+\sqrt{3}}(\sqrt{(2-\sqrt{3})(2+\sqrt{3})}$

$x=2+\sqrt{3}+2\sqrt{2+\sqrt{3}}⟶\left(1\right)$

$\Rightarrow x=2+\sqrt{3}+2\sqrt{2+\sqrt{3}}$

$\therefore \tan \left(82\frac{1}{2}\right)=2+\sqrt{3}+2\sqrt{2+\sqrt{3}}$.