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Question Number 83319 by jagoll last updated on 29/Feb/20

if \tan β = \frac{\tan α + \tan γ}{1 + \tan α \tan γ}

show that \sin 2 β = \frac{\sin 2 α + \sin 2 γ}{1 + \sin 2 α \sin 2 γ}

Answered by mind is power last updated on 01/Mar/20

s i n (2 β) = \frac{2 t g (β)}{1 + {t g}^{2} (β)}

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

s i n (2 β) = 2 \frac{\frac{s i n (a + γ)}{c o s (a - γ)}}{1 + \frac{{s i n}^{2} (a + γ)}{{c o s}^{2} (a - γ)}}

= \frac{2 s i n (a + γ) c o s (a - γ)}{{c o s}^{2} (a - γ) + {s i n}^{2} (a + γ)}

{c o s}^{2} (a - γ) + {s i n}^{2} (a + γ) = {c o s}^{2} (a) {c o s}^{2} (γ) + {s i n}^{2} (a) {s i n}^{2} (γ)

+ {s i n}^{2} (a) {c o s}^{2} (γ) + {s i n}^{2} (γ) {c o s}^{2} (a) + 4 s i n (a) c o s (a) s i n (γ) c o s (γ)

= {c o s}^{2} (γ) ({s i n}^{2} (a) + {c o s}^{2} (a)) + {s i n}^{2} (γ) ({s i n}^{2} (a) + {c o s}^{2} (a))

+ s i n (2 a) s i n (2 γ)

= 1 + s i n (2 γ) c o s (2 a)

2 s i n (a + γ) c o s (a - γ) = s i n (2 γ) + s i n (2 a)

i u s e d, 2 s i n (a) c o s (b) = s i n (a + b) + s i n (a - b)

w e g e t \frac{s i n (2 γ) + s i n (2 a)}{1 + s i n (2 a) s i n (2 γ)}

Commented by peter frank last updated on 01/Mar/20

t h a n k y o u

Commented by jagoll last updated on 01/Mar/20

thank you sir

Commented by mind is power last updated on 01/Mar/20

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