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Geometry Level 3

If cot ⁡ − 1 ( 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x ) = A x \cot^{-1} \left(\dfrac{\sqrt{1+ \sin x} + \sqrt{1- \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right) = Ax , where 0 < x < π 2 0 < x < \dfrac{\pi}{2} , find the value of A A .


The answer is 0.50.

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Samara Simha Reddy by Samara Simha Reddy , 22, India

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3 solutions

Chew-Seong Cheong
May 12, 2016

cot ⁡ A x = 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x = 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x × 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x + 1 − sin ⁡ x = 1 + sin ⁡ x + 2 1 − sin ⁡ 2 x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 + sin ⁡ x = 2 + 2 cos ⁡ x 2 sin ⁡ x = 1 + cos ⁡ x sin ⁡ x Let t = tan ⁡ ( x 2 ) = 1 + 1 − t 2 1 + t 2 2 t 1 + t 2 = 1 + t 2 + 1 − t 2 2 t 2 = 2 2 t = 1 t = cot ⁡ ( x 2 ) \begin{aligned} \cot Ax & = \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \\ & = \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \times \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} \\ & = \frac{1+\sin x + 2\sqrt{1-\sin^2 x} + 1-\sin x}{1+\sin x - 1 + \sin x} \\ & = \frac{2+2\cos x}{2 \sin x} \\ & = \frac{1+\cos x}{\sin x} \quad \quad \small \color{#3D99F6}{\text{Let }t = \tan \left(\frac{x}{2}\right)} \\ & = \frac{1+\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} \\ & = \frac{1+t^2+1-t^2}{2t^2} \\ & = \frac{2}{2t} = \frac{1}{t} = \cot \left(\frac{x}{2}\right) \end{aligned}

⟹ A = 1 2 = 0.5 \implies A = \frac{1}{2} = \boxed{0.5}

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[ ∵ 1 ± sin ⁡ x = cos ⁡ 2 x 2 + sin ⁡ 2 x 2 ± 2 sin ⁡ x 2 cos ⁡ x 2 = ( cos ⁡ x 2 ± sin ⁡ x 2 ) 2 ] \large \displaystyle \left[ \because 1 \pm \sin x = \cos ^{2} \frac{x}{2} + \sin ^{2} \frac{x}{2} \pm 2\sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2} \right)^2 \right]

∴ cot ⁡ − 1 ( 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x ) = cot ⁡ − 1 ( ( cos ⁡ x 2 + sin ⁡ x 2 ) 2 + ( cos ⁡ x 2 − sin ⁡ x 2 ) 2 ( cos ⁡ x 2 + sin ⁡ x 2 ) 2 − ( cos ⁡ x 2 − sin ⁡ x 2 ) 2 ) = cot ⁡ − 1 ( ∣ cos ⁡ x 2 + sin ⁡ x 2 ∣ + ∣ cos ⁡ x 2 − sin ⁡ x 2 ∣ ∣ cos ⁡ x 2 + sin ⁡ x 2 ∣ − ∣ cos ⁡ x 2 − sin ⁡ x 2 ∣ ) = cot ⁡ − 1 ( cos ⁡ x 2 + sin ⁡ x 2 + cos ⁡ x 2 − sin ⁡ x 2 cos ⁡ x 2 + sin ⁡ x 2 − cos ⁡ x 2 + sin ⁡ x 2 ) = cot ⁡ − 1 ( c o t x 2 ) = x 2 \large \displaystyle \begin{aligned} \therefore \cot^{-1} \left(\frac{\sqrt{1+ \sin x} + \sqrt{1- \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right)\\ \large \displaystyle = \cot^{-1} \left(\frac{\sqrt{\left(\cos \frac{x}{2} + \sin \frac{x}{2} \right)^2} + \sqrt{\left(\cos \frac{x}{2} - \sin \frac{x}{2} \right)^2}}{\sqrt{\left(\cos \frac{x}{2} + \sin \frac{x}{2} \right)^2} - \sqrt{\left(\cos \frac{x}{2} - \sin \frac{x}{2} \right)^2}} \right)\\ \large \displaystyle = \cot^{-1} \left(\frac{\left|\cos \frac{x}{2} + \sin \frac{x}{2}\right| + \left|\cos \frac{x}{2} - \sin \frac{x}{2} \right|}{\left|\cos \frac{x}{2} + \sin \frac{x}{2}\right| - \left|\cos \frac{x}{2} - \sin \frac{x}{2} \right|} \right)\\ \large \displaystyle = \cot^{-1} \left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2} - \cos\frac{x}{2} + \sin \frac{x}{2}} \right)\\ \large \displaystyle = \cot^{-1} \left(cot \frac{x}{2} \right) \\ \large \displaystyle = \color{#3D99F6}{\boxed{\frac{x}{2}}} \end{aligned}

⟹ x 2 = A x ⟹ A = 1 2 = 0.5 . \large \displaystyle \implies \frac{x}{2} = Ax \\ \large \displaystyle \implies A = \frac{1}{2} = \color{#69047E}{\boxed{0.5}}.

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Isn't it x 2 = A x \dfrac{x}{2} = Ax ??

Hung Woei Neoh - 5 years, 1 month ago

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Great! Thanks!

Samara Simha Reddy - 5 years, 1 month ago

L e t f ( x ) = ( 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x ) M u l t i p l y i n g n u m e r a t o r a n d d e n o m i n a t o r b y 1 + sin ⁡ x + 1 − sin ⁡ x , a n d s i m p l i f y i n g , f ( x ) = 2 + 2 c o s x 2 s i n x = 2 ∗ c o s 2 ( 1 2 ∗ x ) 2 ∗ s i n ( 1 2 ∗ x ) ∗ c o s ( 1 2 ∗ x ) ∵ 1 + c o s T = 2 ∗ c o s 2 ( 1 2 ∗ T ) a n d s i n T = 2 ∗ s i n ( 1 2 ∗ T ) ∗ c o s ( 1 2 ∗ T ) . ∴ f ( x ) = c o s ( 1 2 ∗ x ) s i n ( 1 2 ∗ x ) = c o t ( 1 2 ∗ x ) . ∴ c o t − 1 f ( x ) = c o t − 1 c o t ( 1 2 ∗ x ) = 12 ∗ x = A ∗ x . ⟹ A = 0.5 Letf(x)= \left(\dfrac{\sqrt{1+ \sin x} + \sqrt{1- \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right)\\ Multiplying\ numerator\ and\ denominator\ by\ \sqrt{1+ \sin x} + \sqrt{1- \sin x},\ and\ simplifying,\\ f(x)=\dfrac {2+2cos x}{2sin x}=\dfrac{2*cos^2(\frac 1 2* x)}{2*sin (\frac 1 2* x)*cos (\frac 1 2* x)} \because\ 1+cos T=2*cos^2(\frac 1 2 *T)\ \ and\ \ sin T=2*sin(\frac 1 2*T)*cos(\frac 1 2 *T).\\ \therefore\ f(x)=\dfrac{cos (\frac 1 2* x)}{sin(\frac 1 2* x)}=cot (\frac 1 2* x).\\ \therefore\ cot^{-1}f(x)=cot^{-1}cot (\frac 1 2* x)=1 2* x=A*x.\\ \implies\ A=\color{#D61F06}{0.5}\ \

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