If-tan2x-sin2x-b-and-tan2x-sin2x-a-prove-that-b-2-a-2-16ba-

Question Number 6655 by Tawakalitu. last updated on 09/Jul/16

I f t a n 2 x - s i n 2 x = b a n d t a n 2 x + s i n 2 x = a

p r o v e t h a t :

b^{2} - a^{2} = 16 b a

Commented by Yozzii last updated on 09/Jul/16

x = \frac{π}{8} \Rightarrow b = 1 - \frac{1}{\sqrt{2}}, a = 1 + \frac{1}{\sqrt{2}}

\Rightarrow 16 a b = 16 (1 - \frac{1}{\sqrt{2}}) (1 + \frac{1}{\sqrt{2}}) = 8

b^{2} - a^{2} = (b - a) (b + a) = (\frac{- 2}{\sqrt{2}}) (2) = - 2 \sqrt{2} \neq 8

b u t {(- 2 \sqrt{2})}^{2} = 8 \Rightarrow {(b^{2} - a^{2})}^{2} = 16 a b ? ? ?

Answered by Yozzii last updated on 09/Jul/16

16 a b = 16 (t a n 2 x - s i n 2 x) (t a n 2 x + s i n 2 x)

= 16 ({t a n}^{2} 2 x - {s i n}^{2} 2 x)

= 16 {s i n}^{2} 2 x (\frac{1}{{c o s}^{2} 2 x} - 1)

= 16 {s i n}^{2} 2 x \frac{1 - {c o s}^{2} 2 x}{{c o s}^{2} 2 x}

= 16 {t a n}^{2} 2 {x s i n}^{2} 2 x

= {(4 t a n 2 x s i n 2 x)}^{2}

= {((2 t a n 2 x) (2 s i n 2 x))}^{2}

= {((t a n 2 x - s i n 2 x + t a n 2 x + s i n 2 x) (t a n 2 x - s i n 2 x - t a n 2 x - s i n 2 x))}^{2}

= {((b + a) (b - a))}^{2}

16 a b = {(b^{2} - a^{2})}^{2}

Commented by Tawakalitu. last updated on 09/Jul/16

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