Trigonometry-If-tan-2-and-tan-2-are-the-roots-of-the-equation-8x-2-26x-15-0-then-what-is-cos-equal-to-

Question Number 134788 by bramlexs22 last updated on 07/Mar/21

Trigonometry

If tan^2(α) and tan^2(β) are the roots of the equation 8x^2−26x+15=0, then what is cos(α+β) equal to?

Answered by john_santu last updated on 07/Mar/21

I f \tan^{2} α a n d \tan^{2} β a r e t h e r o o t s o f

t h e e q u a t i o n 8 x^{2} - 26 x + 15 = 0

t h e n w h a t i s \cos (α + β) e q u a l t o ?

L e t {\begin{cases} \tan α \\ \tan β \end{cases} a r e t h e r o o t s o f e q u a t i o n

8 {(\sqrt{u})}^{2} - 26 (\sqrt{u}) + 15 = 0

o r 64 u^{2} - 436 u + 225 = 0

B y {V i e t a}^{'} s t h e o r e m

{\begin{cases} \tan α + \tan β = \frac{436}{64} \dots (i) \\ \tan α . \tan β = \frac{225}{64} \dots (i i) \end{cases}

s o w e g e t \tan (α + β) = \frac{\frac{436}{64}}{1 - \frac{225}{64}} = - \frac{436}{161}

t h e n \sec^{2} (α + β) = 1 + {(- \frac{436}{161})}^{2} = \frac{216 017}{25 921}

∴ \cos (α + β) = \frac{1}{\sec (α + β)} = - \sqrt{\frac{25 921}{216 017}}

\cos (α + β) \approx - 0.3464

N o t e : w e u s e n e g a t i v e r o o t b e c a u s e

b y e q (i) a n d (i i) α, β a r e a c u t e a n d

α + β i s i n 2^{n d} q u a d r a n t .