Given-tan-and-tan-are-the-two-roots-of-2x-2-x-2-0-then-sin-2-2-cos-2-2-tan-2-2-

Question Number 110254 by ZiYangLee last updated on 28/Aug/20

Given \tan α and \tan β are the two roots

of 2 x^{2} - x - 2 = 0, then

\sin (2 α + 2 β) + \cos (2 α + 2 β) + \tan (2 α + 2 β) = ?

Answered by som(math1967) last updated on 28/Aug/20

\tan α + \tan β = \frac{1}{2}

\tan α \tan β = \frac{- 2}{2} = - 1

\tan (α + β) = \frac{\frac{1}{2}}{1 - (- 1)} = \frac{1}{4}

\sin 2 (α + β) + \cos 2 (α + β) + \tan 2 (α + β)

\frac{2 \tan (α + β)}{1 + \tan^{2} (α + β)} + \frac{1 - \tan^{2} (α + β)}{1 + \tan^{2} (α + β)}

+ \frac{2 \tan (α + β)}{1 - \tan^{2} (α + β)}

now put \tan (α + β) = \frac{1}{4}

Commented by ZiYangLee last updated on 28/Aug/20

Thanks!