asin-bcos-c-acosec-bsec-c-prove-that-sin2-2ab-c-2-a-2-b-2-

Question Number 102390 by Dwaipayan Shikari last updated on 08/Jul/20

a s i n θ + b c o s θ = c

a c o s e c θ + b s e c θ = c

p r o v e t h a t s i n 2 θ = \frac{2 a b}{c^{2} - a^{2} - b^{2}}

Commented by PRITHWISH SEN 2 last updated on 08/Jul/20

Just multiply the two equations

Commented by bobhans last updated on 09/Jul/20

i f a^{2} + b^{2} ⩾ c^{2} o r a^{2} + b^{2} ⩽ c^{2} ?

Commented by bemath last updated on 09/Jul/20

(2) \frac{a}{\sin θ} + \frac{b}{\cos θ} = c

\frac{a \cos θ + b \sin θ}{\frac{1}{2} \sin 2 θ} = c \Rightarrow a \cos θ + b \sin θ = \frac{1}{2} c \sin 2 θ

(1) \times (2)

(a \sin θ + b \cos θ) (a \cos θ + b \sin θ) = \frac{1}{2} c^{2} \sin 2 θ

\frac{1}{2} a^{2} \sin 2 θ + a b \sin^{2} θ + a b \cos^{2} θ + \frac{1}{2} b^{2} \sin 2 θ = \frac{1}{2} c^{2} \sin 2 θ

{\frac{1}{2} a^{2} + \frac{1}{2} b^{2}} \sin 2 θ + a b = \frac{1}{2} c^{2} \sin 2 θ

a b = {\frac{1}{2} c^{2} - \frac{1}{2} a^{2} - \frac{1}{2} b^{2}} \sin 2 θ

\sin 2 θ = \frac{2 a b}{c^{2} - a^{2} - b^{2}}

Commented by 1549442205 last updated on 09/Jul/20

Answered by Dwaipayan Shikari last updated on 08/Jul/20

(a s i n θ + b c o s θ) (a c o s θ + b s i n θ) . \frac{2}{s i n 2 θ} = c^{2}

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

\frac{s i n 2 θ}{s i n 2 θ} (a^{2} + b^{2}) + \frac{2 a b}{s i n 2 θ} = c^{2}

s i n 2 θ = \frac{2 a b}{c^{2} - a^{2} - b^{2}} (p r o v e d)

Commented by bobhans last updated on 09/Jul/20

I f {\begin{cases} 2 \sin θ + \cos θ = 2 \\ 2 \csc θ + \sec θ = 2 \end{cases}

\Leftrightarrow \sin 2 θ = \frac{2.2.1}{4 - 4 - 1} = - 4 ? ?

i t c o r r e c t ?

Commented by 1549442205 last updated on 09/Jul/20

but the your system has no roots!

the first eqs implies that \sin 2 θ = \frac{24}{25}!

Commented by PRITHWISH SEN 2 last updated on 09/Jul/20

a, b, and c cannot be an arbitary choice for these

identities . It should follow the certain condition

that

- 1 ⩽ \frac{2 ab}{c^{2} - a^{2} - b^{2}} ⩽ 1

such pair can be

a = 1 b = 2 and c = 5 and so on .

Commented by bemath last updated on 09/Jul/20

i t d o e s m e a n t c^{2} ⩾ a^{2} + b^{2}

Answered by 1549442205 last updated on 09/Jul/20

(1) \Leftrightarrow \frac{a}{\sqrt{a^{2} + b^{2}}} \sin θ + \frac{b}{\sqrt{a^{2} + b^{2}}} \cos θ = \frac{c}{\sqrt{a^{2} + b^{2}}}

\Leftrightarrow \sin (φ + θ) = \sin [\sin^{- 1} (\frac{c}{\sqrt{a^{2} + b^{2}}})] (where φ = \arccos \frac{a}{\sqrt{a^{2} + b^{2}}})

(with the condition a^{2} + b^{2} ⩾ c^{2})

\Rightarrow φ + θ = \sin^{- 1} (\frac{c}{\sqrt{a^{2} + b^{2}}}) + 2 k π

\Rightarrow θ = \sin^{- 1} (\frac{c}{\sqrt{a^{2} + b^{2}}}) + 2 k π - φ

\Rightarrow \sin θ = \sin (\sin^{- 1} (\frac{c}{\sqrt{a^{2} + b^{2}}}) - φ)

= \frac{c}{\sqrt{a^{2} + b^{2}}} \cos φ - \frac{\sqrt{a^{2} + b^{2} - c^{2}}}{\sqrt{a^{2} + b^{2}}} \sin φ

= \frac{c}{\sqrt{a^{2} + b^{2}}} . \frac{a}{\sqrt{a^{2} + b^{2}}} - \frac{\sqrt{a^{2} + b^{2} - c^{2}}}{\sqrt{a^{2} + b^{2}}} . \frac{b}{\sqrt{a^{2} + b^{2}}}

= \frac{ac - b \sqrt{a^{2} + b^{2} - c^{2}}}{a^{2} + b^{2}}

\Rightarrow \cos θ = \frac{1}{a^{2} + b^{2}} \sqrt{{(a^{2} + b^{2})}^{2} - (ac - b \sqrt{{a^{2} + b^{2} - c^{2})}^{2}}}

\Rightarrow \sin 2 θ = 2 \sin θ \cos θ = 2 (\frac{ac - b \sqrt{a^{2} + b^{2} - c^{2}}}{a^{2} + b^{2}}) (\frac{1}{a^{2} + b^{2}} \sqrt{{(a^{2} + b^{2})}^{2} - (ac - b \sqrt{{a^{2} + b^{2} - c^{2})}^{2}}}) = f (a, b, c)