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Question Number 127169 by bramlexs22 last updated on 27/Dec/20

I f (1 + \cos x) (1 + \sin x) = \frac{3}{2}

t h e n (1 - \cos x) (1 - \sin x) = ?

n i c e t r i g o n o m e t r y

Commented by Dwaipayan Shikari last updated on 27/Dec/20

(1 - c o s x) (1 - s i n x) (1 + c o s x) (1 + s i n x) = \frac{3}{2} (1 - c o s x) (1 - s i n x)

\Rightarrow {s i n}^{2} x {c o s}^{2} x = \frac{3}{2} (1 - c o s x) (1 - s i n x) = \frac{3}{2} a

1 + c o s x + s i n x + s i n x c o s x = \frac{3}{2} \to (a)

1 - c o s x - s i n x + s i n x c o s x = a \to (b)

(a + b) \to 2 s i n x c o s x = a - \frac{1}{2}

\frac{3}{2} a = \frac{1}{4} {(a - \frac{1}{2})}^{2} \Rightarrow a^{2} + \frac{1}{4} - a = 6 a \Rightarrow a^{2} - 7 a + \frac{1}{4} \Rightarrow a = \frac{7}{2} \pm 2 \sqrt{3}

H e r e a = \frac{7}{2} - 2 \sqrt{3}

Commented by bramlexs22 last updated on 27/Dec/20

Commented by liberty last updated on 27/Dec/20

(1 + \sin x) (1 + \cos x) = \frac{3}{2}

{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2} (2 \cos^{2} \frac{x}{2}) = \frac{3}{2}

(\sin \frac{x}{2} + \cos \frac{x}{2}) . \cos \frac{x}{2} = \pm \frac{\sqrt{3}}{2}

\sin \frac{x}{2} . \cos \frac{x}{2} + \cos^{2} \frac{x}{2} = \pm \frac{\sqrt{3}}{2}

\frac{\sin x}{2} + \frac{1 + \cos x}{2} = \pm \frac{\sqrt{3}}{2}

\Rightarrow \sin x + \cos x = \sqrt{3} - 1; o t h e r v a l u e i s

o u t s i d e t h e r a n g e .

n o t e : r a n g e o f (\sin x + \cos x) = [- \sqrt{2}, \sqrt{2}]

a n d - 1 - \sqrt{3} < - \sqrt{2}

∴ {(\sin x + \cos x)}^{2} = {(\sqrt{3} - 1)}^{2}

1 + 2 \sin x \cos x = 4 - 2 \sqrt{3}

t h e n f (x) = (1 - \sin x) (1 - \cos x)

= 1 - (\sin x + \cos x) + \sin x \cos x

= 1 - (\sqrt{3} - 1) + \frac{3 - 2 \sqrt{3}}{2}

= \frac{7 - 4 \sqrt{3}}{2} \approx 0.035898

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