﻿ Find the value of \dfrac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}

#### Question

Find the value of \dfrac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}

Collected in the board: Trigonometry

Steven Zheng posted 18 hours ago

Step 1 - Using Sum to product identities on both denominator and numerator yieds the value of the following trigonometric expression.

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

Step 2 - Use the half angle identities, show that

\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} }

Apply the half angle identities to both denumerator and numerator.

\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} }

Step 3 - Apply the trigonometric result to the square expressions.

\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{99}}

=\sqrt{10}(\sqrt{1+\sqrt{0.01}}+\sqrt{1+\sqrt{0.99}})

=\sqrt{10}(\sqrt{1+\sin(t)}+\sqrt{1+\cos(t)}

=\sqrt{10}(1+\sqrt 2)(\sqrt{1-\sin(t)}+\sqrt{1-\cos(t)})

=\sqrt{10}(1+\sqrt 2)(\sqrt{1-\sqrt{0.01}}+\sqrt{1-\sqrt{0.99}})

=(1+\sqrt 2)(\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{99}})

Similarly, we get

\sqrt{10+\sqrt{2}}+\sqrt{10+\sqrt{98}} = (\sqrt{2}+1)(\sqrt{10-\sqrt{2}}+\sqrt{10-\sqrt{98}} )

\dots

\sqrt{10+\sqrt{49}}+\sqrt{10+\sqrt{51}} = (\sqrt{2}+1)(\sqrt{10-\sqrt{49}}+\sqrt{10-\sqrt{51}} )

For a special case

\dfrac{\sqrt{10+\sqrt{50} } }{\sqrt{10-\sqrt{50} } } = \dfrac{\sqrt{2+\sqrt{2} } }{\sqrt{2-\sqrt{2} } } = \sqrt{2}+1

Then,

\sqrt{10+\sqrt{50} } = (\sqrt{2}+1)\sqrt{10-\sqrt{50} }