If-sin-sin-2-sin-3-cos-and-0-lt-lt-pi-2-then-find-

Question Number 140477 by ajfour last updated on 08/May/21

I f \sin θ + \sin^{2} θ + \sin^{3} θ + \dots .

= \cos θ a n d 0 < θ < \frac{π}{2} t h e n

f i n d θ .

Answered by benjo_mathlover last updated on 08/May/21

\Rightarrow \frac{\sin θ}{1 - \sin θ} = \cos θ

let \sin θ = x \Rightarrow \cos θ = \sqrt{1 - x^{2}}

\Rightarrow \frac{x}{1 - x} = \sqrt{1 - x^{2}}

\Rightarrow x \approx 0.468989

\Rightarrow θ = \sin^{- 1} (0.468989)

\sin^{- 1} (0.468989) = 0.488146 rad

Commented by ajfour last updated on 08/May/21

Thanks sir, i also tried.

Answered by ajfour last updated on 08/May/21

\frac{\sin θ}{1 - \sin θ} = \cos θ

p^{2} + q^{2} = 1

\frac{p}{1 - p} = q

p q = t

t^{2} + q^{4} = q^{2}

t = q^{2} - q t

q^{2} = q t + t

(q^{2} t^{2} + t^{2} + 2 {q t}^{2}) = q t + t - t^{2}

{q t}^{3} + t^{3} + t^{2} + 2 {q t}^{2} = q t + t - t^{2}

\Rightarrow q (t^{3} + 2 t^{2} - t) + (t^{3} + 2 t^{2} - t) = 0

l e t t^{2} + 2 t - 1 = 0

\Rightarrow {(t + 1)}^{2} = 2

t = - 1 + \sqrt{2}

\Rightarrow \sin θ \cos θ = \sqrt{2} - 1

\sin 2 θ = 2 \sqrt{2} - 2

θ = \frac{1}{2} \sin^{- 1} (2 \sqrt{2} - 2)

θ \approx 0.488146802 r a d

Answered by mr W last updated on 08/May/21

\sin θ + \sin^{2} θ + \sin^{3} θ + \dots = \cos θ

\sin θ + \sin^{2} θ + \sin^{3} θ + \dots

= \sin θ (1 + \sin θ + \sin^{2} θ + \sin^{3} θ + \dots)

= \sin θ \times (1 + \cos θ)

\sin θ + \sin^{2} θ + \sin^{3} θ + \dots

= \sin θ \times \frac{1}{1 - \sin θ}

\Rightarrow 1 + \cos θ = \frac{1}{1 - \sin θ}

\Rightarrow \cos θ - \sin θ = \sin θ \sin θ

\Rightarrow 1 - 2 \sin θ \cos θ = {(\sin θ \sin θ)}^{2}

\Rightarrow {(\sin θ \cos θ)}^{2} + 2 \sin θ \cos θ - 1 = 0

\Rightarrow \sin θ \cos θ = \sqrt{2} - 1

\Rightarrow 2 \sin θ \cos θ = 2 (\sqrt{2} - 1)

\Rightarrow \sin 2 θ = 2 (\sqrt{2} - 1)

\Rightarrow θ = \frac{\sin^{- 1} 2 (\sqrt{2} - 1)}{2}

Commented by ajfour last updated on 08/May/21

n i c e s i r, y o u g o t i t s t r a i g h t!

Facebook Tweet Pin