If${\ tan}{\alpha/2\ and an{\beta/2\ \ }}$are the roots of the equation $8x^{2} - \ 26x + 15 = 0$ then $\cos(\alpha + \beta)$ is equal to

Question

If${\ tan}{\alpha/2\ and	an{\beta/2\ \ }}$are the roots of the equation $8x^{2} - \ 26x + 15 = 0$ then $\cos(\alpha + \beta)$ is equal to

Dr. Abubacker · Accepted Answer

(a)
$\because	an{\alpha/2\  +}	an{\beta/2\ } = \ \frac{26}{8} = \frac{13}{4}$
and $	an{\alpha/2}	an{\beta/2 = \frac{15}{8}}$
$	herefore	an{\left( \frac{\alpha + \beta}{2} ight) = \frac{	an{\alpha/2} + 	an{\beta/2}}{1 - 	an{\alpha/2}	an{\beta/2}}}$
$= \frac{\frac{13}{4}}{1 - \frac{15}{8}} = - \frac{26}{7}$
Now, $\cos{(\alpha + \beta) = \frac{1 - 	an^{2}\left( \frac{\alpha + \beta}{2} ight)}{1 + 	an^{2}\left( \frac{\alpha + \beta}{2} ight)}}$
$= \frac{1 - \left( - \frac{26}{7} ight)^{2}}{1 + \left( - \frac{26}{7} ight)^{2}} = \frac{49 - 676}{49 + 676}$
$= - \frac{627}{725}$

Q. If${\ tan}{\alpha/2\ and\tan{\beta/2\ \ }}$are the roots of the equation $8x^{2} - \ 26x + 15 = 0$ then $\cos(\alpha + \beta)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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