﻿ Show that \dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x} }{\sqrt{1-\sin x}+\sqrt{1-\cos x} } =\sqrt{2}+1

#### Question

Show that \dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x} }{\sqrt{1-\sin x}+\sqrt{1-\cos x} } =\sqrt{2}+1

Collected in the board: Trigonometry

Steven Zheng posted 20 hours ago

The first step is to show that \dfrac{\cos x+\cos\Big( \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) } =\sqrt{2}+1

Using Sum to product identities

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

Here

\alpha = x, \beta = \dfrac{\pi}{4}-x

\dfrac{\alpha +\beta }{2} = \dfrac{\pi}{8} , \dfrac{\alpha - \beta }{2} = x-\dfrac{\pi}{8}

\dfrac{\cos x+\cos\Big( \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }

=\dfrac{2\cos\dfrac{\pi}{8}\cos(x-\dfrac{\pi}{8} ) }{2\sin\dfrac{\pi}{8}\cos(x-\dfrac{\pi}{8} ) } = \cot\dfrac{\pi}{8}

Using half angle identity for tangent function to find the value of \tan\dfrac{\pi}{8} ,

\tan \dfrac{\alpha }{2}=\sqrt{\dfrac{1-\cos \alpha }{1+\cos \alpha} }

\tan\dfrac{\pi}{8}=\sqrt{\dfrac{1-\cos\dfrac{\pi}{4} }{1+\cos\dfrac{\pi}{4} } }

=\sqrt{\dfrac{1-\dfrac{\sqrt{2} }{2} }{1+\dfrac{\sqrt{2} }{2} } }

=\sqrt{\dfrac{2-\sqrt{2} }{2+\sqrt{2} } }

=\dfrac{2-\sqrt{2} }{\sqrt{2} } = \sqrt{2}-1

Then, the value of \cot\dfrac{\pi}{8} is the reciprocal of \tan\dfrac{\pi}{8}

\cot\dfrac{\pi}{8} = \dfrac{1}{\tan\dfrac{\pi}{8} }

=\dfrac{1}{\sqrt{2} -1 } =\sqrt{2}+1

Therefore,

\dfrac{\cos x+\cos\Big( \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }=\sqrt{2}+1

Next step is to show that

which is silver ratio.

\dfrac{\cos x+\cos\Big( \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) } =\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x} }{\sqrt{1-\sin x}+\sqrt{1-\cos x} }

Use the half angle identities

\sin \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1-\cos \alpha }{2} }

\cos \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1+\cos \alpha }{2} }

\dfrac{\cos x+\cos\Big( \dfrac{\pi}{4}-x \Big) }{\sin x+\sin\Big( \dfrac{\pi}{4}-x\Big) }

=\dfrac{\sqrt{\dfrac{1+\cos x }{2} }+ \sqrt{\dfrac{1+\cos (\dfrac{\pi}{2} - x) }{2}}}{\sqrt{\dfrac{1-\cos x }{2} }+ \sqrt{\dfrac{1-\cos (\dfrac{\pi}{2} - x) }{2}}}

=\dfrac{\sqrt{1+\sin x}+\sqrt{1+\cos x} }{\sqrt{1-\sin x}+\sqrt{1-\cos x} }

Therefore

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

Steven Zheng posted 21 minutes ago

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