when-tan-2-1-a-then-find-cos-from-the-a-

Question Number 193423 by mustafazaheen last updated on 13/Jun/23

when \tan \frac{θ}{2} = \frac{1}{a}

then find \cos θ = ? from the a

Answered by AST last updated on 13/Jun/23

t a n (\frac{θ}{2} + \frac{θ}{2}) = \frac{2 t a n (\frac{θ}{2})}{1 - {t a n}^{2} (\frac{θ}{2})} \Rightarrow t a n (θ) = \frac{2 a}{a^{2} - 1}

{s e c}^{2} (θ) = 1 + {t a n}^{2} θ = 1 + \frac{4 a^{2}}{{(a^{2} - 1)}^{2}} = \frac{{(a^{2} - 1)}^{2} + 4 a^{2}}{{(a^{2} - 1)}^{2}}

\Rightarrow c o s θ = \frac{a^{2} - 1}{\sqrt{{(a^{2} + 1)}^{2}}} = \frac{a^{2} - 1}{a^{2} + 1} = 1 - \frac{2}{a^{2} + 1}

Commented by mustafazaheen last updated on 13/Jun/23

\underset{⏟}{T}

Answered by Subhi last updated on 13/Jun/23

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

s i n (\frac{θ}{2}) = \sqrt{\frac{1 - c o s (θ)}{2}}

c o s (\frac{θ}{2}) = \sqrt{\frac{1 + c o s (θ)}{2}}

\frac{s i n (\frac{θ}{2})}{c o s (\frac{θ}{2})} = t a n (\frac{θ}{2}) = \frac{1}{a} = \sqrt{\frac{1 - c o s (θ)}{1 + c o s (θ)}}

a^{2} (1 - c o s (θ)) = 1 + c o s (θ) ⇛ a^{2} - 1 = (1 + a^{2}) c o s (θ)

c o s (θ) = \frac{a^{2} - 1}{a^{2} + 1}

Commented by mustafazaheen last updated on 13/Jun/23

⋚

Commented by York12 last updated on 16/Jun/23

⋚

Answered by BaliramKumar last updated on 13/Jun/23

\cos θ = \frac{1 - \tan^{2} (\frac{θ}{2})}{1 + \tan^{2} (\frac{θ}{2})} = \frac{1 - {(\frac{1}{a})}^{2}}{1 + {(\frac{1}{a})}^{2}} = \frac{1 - \frac{1}{a^{2}}}{1 + \frac{1}{a^{2}}}

\cos θ = \frac{a^{2} - 1}{a^{2} + 1}

pythagorean triplet \Rightarrow \underset{P}{\underset{⏟}{2 a}}, \underset{B}{\underset{⏟}{(a^{2} - 1)}}, \underset{H}{\underset{⏟}{(a^{2} + 1)}}

\tan θ = \frac{2 a}{a^{2} - 1}