Prove-that-cot7-5-2-3-4-6-

Question Number 102587 by 1549442205 last updated on 10/Jul/20

Prove that

\cot 7 . 5 ° = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}

Commented by PRITHWISH SEN 2 last updated on 10/Jul/20

a different approach

\cot θ = \frac{\cos θ}{\sin θ} = \frac{2 \cos 7 θ \cos θ}{2 \cos 7 θ \sin θ}

= \frac{\cos 8 θ + \cos 6 θ}{\sin 8 θ - \sin 6 θ}

put θ = 7 \cdot 5

\cot 7 \cdot 5 = \frac{\cos 60 + \cos 45}{\sin 60 - \sin 45} = \frac{(1 + \sqrt{2})}{(\sqrt{3} - \sqrt{2})} = (1 + \sqrt{2}) (\sqrt{3} + \sqrt{2})

= \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}

Answered by Dwaipayan Shikari last updated on 10/Jul/20

M e t h o d 1) c o t 7.5 ° = \frac{c o s 7.5 °}{s i n 7.5} = \frac{2 {c o s}^{2} 7.5 °}{s i n 15 °} = \frac{1 + c o s 15 °}{s i n 15 °} = \frac{2 c o s 15 ° + 1 + c o s 30 °}{s i n 30 °}

c o s 15 = \frac{\sqrt{3} + 1}{2 \sqrt{2}} = 2 (\frac{\sqrt{3} + 1}{\sqrt{2}} + 1 + \sqrt{3} . \frac{1}{2})

= 2 (\frac{\sqrt{6} + \sqrt{2} + \sqrt{4} + \sqrt{3}}{2})

s o c o t 7.5 ° = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} (p r o v e d)

M e t h o d 2

c o s \frac{π}{24} = \sqrt{\frac{1}{2} (1 + c o s \frac{π}{12})} = \sqrt{\frac{1}{2} (1 + \frac{\sqrt{3} + 1}{2 \sqrt{2}})} = \frac{1}{2} \sqrt{\frac{2 \sqrt{2} + \sqrt{3} + 1}{\sqrt{2}}}

\frac{1}{2 \sqrt{2}} \sqrt{4 + \sqrt{6} + \sqrt{2}}

s i n \frac{π}{24} = \frac{1}{2 \sqrt{2}} . \sqrt{4 - \sqrt{6} - \sqrt{2}}

c o t \frac{π}{24} = \sqrt{\frac{4 + \sqrt{6} + \sqrt{2}}{4 - \sqrt{6} - \sqrt{2}}} = \frac{4 + \sqrt{6} + \sqrt{2}}{16 - 8 - 2 \sqrt{12}} = \frac{1}{2} . \frac{4 + \sqrt{6} + \sqrt{2}}{4 - \sqrt{12}} = \frac{1}{2} . \frac{(4 + \sqrt{6} + \sqrt{2}) (4 + \sqrt{12})}{4}

= \frac{1}{8} (16 + 4 \sqrt{6} + 4 \sqrt{2} + 4 \sqrt{12} + 6 \sqrt{2} + 2 \sqrt{12})

= 2 + \sqrt{\frac{3}{2}} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{3}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}}

= \sqrt{4} + \frac{2 \sqrt{3} + 2 + 2 \sqrt{6} + 3 + \sqrt{3}}{2 \sqrt{2}} = \sqrt{4} + \sqrt{2} + \sqrt{3} + \sqrt{6}

Commented by bemath last updated on 10/Jul/20

h o w \frac{\cos 7 . 5^{o}}{\sin 7 . 5^{o}} = \frac{2 \cos^{2} 7 . 5^{o}}{\sin 7 . 5^{o}} ?

Commented by Dwaipayan Shikari last updated on 10/Jul/20

m u l t i p l y i n g b y 2 c o s 7.5 °

Commented by Dwaipayan Shikari last updated on 10/Jul/20

I t i s s i n 15 °

Commented by bemath last updated on 10/Jul/20

o o \sin 15^{o} . t h a n k

Answered by bobhans last updated on 10/Jul/20

\tan 15^{o} = \tan (2 \times 7 . 5^{o}) = \frac{2 \tan 7 . 5^{o}}{1 - \tan^{2} (7 . 5^{o})}

(1) \tan 15^{o} = \tan (45^{o} - 30^{o}) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} =

\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}

(2) s e t \tan 7 . 5^{o} = x

(2 - \sqrt{3}) (1 - x^{2}) = 2 x

(2 - \sqrt{3}) x^{2} + 2 x - (2 - \sqrt{3}) = 0

x = \frac{- 2 + \sqrt{4 + 4 {(2 - \sqrt{3})}^{2}}}{2 (2 - \sqrt{3})} =

\frac{- 1 + \sqrt{1 + 7 - 4 \sqrt{3}}}{2 - \sqrt{3}} = \frac{- 1 + \sqrt{8 - 2 \sqrt{12}}}{2 - \sqrt{3}} =

\frac{- 1 + \sqrt{6} - \sqrt{2}}{2 - \sqrt{3}} = \frac{(\sqrt{6} - \sqrt{2} - 1) (2 + \sqrt{3})}{1} =

2 \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{2} - \sqrt{6} - 2 - \sqrt{3} =

\sqrt{6} + \sqrt{2} - 2 - \sqrt{3} = \tan 7 . 5^{o}

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