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Question Number 109337 by ZiYangLee last updated on 22/Aug/20

x = \cos θ, where \frac{3 π}{2} < θ < 2 π, and that 2 \cos θ - \sin θ = 2,

show that \sqrt{1 - x^{2}} = 2 (1 - x) .

Hence or otherwise, find x and deduce that \tan 2 θ = \frac{24}{7}

Answered by Aziztisffola last updated on 22/Aug/20

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

= 2 - 2 \cos θ = 2 (1 - \cos θ) = 2 (1 - x)

\sqrt{1 - x^{2}} = 2 (1 - x) \Leftrightarrow 1 - x^{2} = 4 (x^{2} - 2 x + 1)

5 x^{2} - 8 x + 3 = 0

△ = 4 \Rightarrow x_{1} = 1 \land x_{2} = \frac{3}{5}

θ \neq 2 π \Rightarrow x \neq 1 \Rightarrow x = \frac{3}{5}

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

\tan θ = \frac{- \frac{4}{5}}{\frac{3}{5}} = - \frac{4}{3}

\tan 2 θ = \frac{2 \times \frac{- 4}{3}}{1 - {(\frac{- 4}{3})}^{2}} = \frac{\frac{- 8}{3}}{1 - \frac{16}{9}}

= \frac{8 \times 3}{7} = \frac{24}{7}

Commented by ZiYangLee last updated on 23/Aug/20

NICE!

Answered by Don08q last updated on 22/Aug/20

x = \cos θ

x^{2} = \cos^{2} θ

1 - x^{2} = \sin^{2} θ

\pm \sqrt{1 - x^{2}} = \sin θ

B u t f o r t h e i n t e r v a l, \frac{3 π}{2} < θ < 2 π sine

h a s a n e g a t i v e v a l u e .

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

s o \sqrt{1 - x^{2}} = - \sin θ \dots \dots \dots \dots (1)

B u t, 2 \cos θ - \sin θ = 2

\Rightarrow 2 - 2 x = - \sin θ

2 (1 - x) = - \sin θ \dots \dots \dots . . (2)

f r o m (1) a n d (2),

\sqrt{1 - x^{2}} = 2 (1 - x) q e d

I t f o l l o w s t h a t

1 - x^{2} = 4 (1 - 2 x + x^{2})

5 x^{2} - 8 x + 3 = 0

x = 1 o r x = \frac{3}{5}

x cannot be 1, since θ < 2 π

∴ x = \frac{3}{5}

\tan θ = \frac{\sin θ}{\cos θ}

\tan θ = \frac{- 2 (1 - x)}{x}

\tan θ = \frac{2 (x - 1)}{x}

\tan θ = \frac{2 (\frac{3}{5} - 1)}{\frac{3}{5}}

\tan θ = - \frac{4}{3}

B u t \tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}

\tan 2 θ = \frac{2 (- \frac{4}{3})}{1 - {(- \frac{4}{3})}^{2}}

∴ \tan 2 θ = \frac{24}{7} q e d

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