calculate-cos-6-pi-8-cos-6-3pi-8-cos-6-5pi-8-cos-6-7pi-8-

Question Number 76633 by mathmax by abdo last updated on 28/Dec/19

c a l c u l a t e {c o s}^{6} (\frac{π}{8}) + {c o s}^{6} (\frac{3 π}{8}) + {c o s}^{6} (\frac{5 π}{8}) + {c o s}^{6} (\frac{7 π}{8})

Answered by MJS last updated on 28/Dec/19

\cos \frac{π}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}

\cos \frac{3 π}{8} = \frac{\sqrt{2 - \sqrt{2}}}{2}

\cos \frac{5 π}{8} = - \frac{\sqrt{2 - \sqrt{2}}}{2}

\cos \frac{7 π}{8} = - \frac{\sqrt{2 + \sqrt{2}}}{2}

\Rightarrow answer is 2 \times ({(\frac{\sqrt{2 + \sqrt{2}}}{2})}^{6} + {(\frac{\sqrt{2 + \sqrt{2}}}{2})}^{6}) = \frac{5}{4}

Commented by mathmax by abdo last updated on 29/Dec/19

t h a n k y o u s i r .

Answered by $@ty@m123 last updated on 29/Dec/19

= {c o s}^{6} (\frac{π}{8}) + {c o s}^{6} (\frac{3 π}{8}) + {c o s}^{6} (\frac{3 π}{8}) + {c o s}^{6} (\frac{π}{8})

= 2 {{c o s}^{6} (\frac{π}{8}) + {c o s}^{6} (\frac{3 π}{8})}

= 2 {{c o s}^{6} (\frac{π}{8}) + {s i n}^{6} (\frac{π}{8})}

= 2 {{c o s}^{2} (\frac{π}{8}) + {s i n}^{2} (\frac{π}{8})} {{c o s}^{4} (\frac{π}{8}) + {s i n}^{4} (\frac{π}{8}) - {c o s}^{2} (\frac{π}{8}) . {s i n}^{2} (\frac{π}{8})}

= 2 {{c o s}^{4} (\frac{π}{8}) + {s i n}^{4} (\frac{π}{8}) - \frac{1}{4} \times 4 {c o s}^{2} (\frac{π}{8}) . {s i n}^{2} (\frac{π}{8})}

= 2 [{{c o s}^{2} (\frac{π}{8}) + {s i n}^{2} (\frac{π}{8})}^{2} - 2 {c o s}^{2} (\frac{π}{8}) . {s i n}^{2} (\frac{π}{8}) - \frac{1}{4} \times {(\sin \frac{π}{4})}^{2}]

= 2 [1 - \frac{1}{2} \times \sin^{2} \frac{π}{4} - \frac{1}{4} \times \frac{1}{2}]

= 2 [1 - \frac{1}{2} \times \frac{1}{2} - \frac{1}{8}]

= 2 (1 - \frac{1}{4} - \frac{1}{8})

= 2 \times \frac{8 - 2 - 1}{8}

= \frac{5}{4}

Commented by abdomathmax last updated on 29/Dec/19

t h a n k x s i r .