if-cos-2-m-2-1-3-tan-3-2-tan-a-prove-that-cos-2-a-1-3-sin-2-a-1-3-2-m-2-1-3-

Question Number 73665 by aliesam last updated on 14/Nov/19

i f

{c o s}^{2} (θ) = \frac{m^{2} - 1}{3}, {t a n}^{3} (\frac{θ}{2}) = t a n (a)

p r o v e t h a t

\sqrt[3]{{c o s}^{2} (a)} + \sqrt[3]{{s i n}^{2} (a)} = \sqrt[3]{{(\frac{2}{m})}^{2}}

Answered by Tanmay chaudhury last updated on 14/Nov/19

c o s θ = \sqrt{\frac{m^{2} - 1}{3}} \to \frac{1 - {t a n}^{2} \frac{θ}{2}}{1 + {t a n}^{2} \frac{θ}{2}} = \sqrt{\frac{m^{2} - 1}{3}} = \frac{a}{1}

\frac{1 + {t a n}^{2} \frac{θ}{2}}{1 - {t a n}^{2} \frac{θ}{2}} = \frac{1}{a} \to \frac{2}{2 {t a n}^{2} \frac{θ}{2}} = \frac{1 + a}{1 - a} \to {t a n}^{2} \frac{θ}{2} = \frac{1 - a}{1 + a}

t a n \frac{θ}{2} = \sqrt{\frac{1 - a}{1 + a} [s o} t a n α = {t a n}^{3} (\frac{θ}{2}) = {(\frac{1 - a}{1 + a})}^{\frac{3}{2}}

{c o s}^{2} α = \frac{1}{1 + {t a n}^{2} α} = \frac{1}{1 + {(\frac{1 - a}{1 + a})}^{3}} = \frac{{(1 + a)}^{3}}{1 + 3 a + 3 a^{2} + a^{3} + 1 - 3 a + 3 a^{2} - a^{3}}

{c o s}^{2} α = \frac{{(1 + a)}^{3}}{2 + 6 a^{2}} \to {s i n}^{2} α = \frac{2 + 6 a^{2} - 1 - 3 a - 3 a^{2} - a^{3}}{2 + 6 a^{2}}

{s i n}^{2} α = \frac{1 - 3 a + 3 a^{2} - a^{3}}{2 + 6 a^{2}} = \frac{{(1 - a)}^{3}}{2 + 6 a^{2}}

{({c o s}^{2} α)}^{\frac{1}{3}} + {({s i n}^{2} α)}^{\frac{1}{3}}

= \frac{1 + a + 1 - a}{{(2 + 6 a^{2})}^{\frac{1}{3}}} = \frac{2}{{{2 + 6 \times \frac{m^{2} - 1}{3})}^{\frac{1}{3}}} = \frac{2}{{(2 + 2 m^{2} - 2)}^{\frac{1}{3}}}

= \frac{2^{1 - \frac{1}{3}}}{m^{\frac{2}{3}}} = {(\frac{2}{m})}^{\frac{2}{3}} p r o v e d