If-sin-23pi-24-2-p-q-1-4-r-then-the-value-of-p-2-q-2-r-2-is-equal-to-

Question Number 133142 by bemath last updated on 19/Feb/21

If \sin (\frac{23 π}{24}) = \sqrt{\frac{2 \sqrt{p} - \sqrt{q} - 1}{4 \sqrt{r}}}

then the value of (p^{2} + q^{2} - r^{2}) is

equal to

Answered by Ñï= last updated on 19/Feb/21

s i n \frac{π}{24}

= \frac{1}{\sqrt{2}} \sqrt{2 {s i n}^{2} \frac{π}{24}}

= \frac{1}{\sqrt{2}} \sqrt{1 - (1 - 2 {s i n}^{2} \frac{π}{24})}

= \frac{1}{\sqrt{2}} \sqrt{1 - c o s \frac{π}{12}}

= \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \sqrt{2 {c o s}^{2} \frac{π}{12} - 1 + 1}}

= \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \sqrt{c o s \frac{π}{6} + 1}}

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

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

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

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

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

p = \frac{1}{2} \dots q = 3 \dots r = \frac{1}{2}

p^{2} + q^{2} - r^{2} = 9

Answered by Dwaipayan Shikari last updated on 19/Feb/21

s i n (\frac{23 π}{24}) = s i n (\frac{π}{24})

1 + c o s \frac{π}{6} = 2 {c o s}^{2} \frac{π}{12}

\Rightarrow \frac{1}{2} \sqrt{2 + \sqrt{3}} = c o s \frac{π}{12}

\Rightarrow 1 - \frac{1}{2} \sqrt{2 + \sqrt{3}} = 2 {s i n}^{2} \frac{π}{24}

\Rightarrow s i n \frac{π}{24} = \frac{1}{2} \sqrt{2 - \sqrt{2 + \sqrt{3}}} = \frac{1}{2} \sqrt{\frac{2 \sqrt{2} - \sqrt{3} - 1}{\sqrt{2}}}

= \sqrt{\frac{2 \sqrt{2} - \sqrt{3} - 1}{4 \sqrt{2}}} p^{2} + q^{2} - r^{2} = 9

A s (\sqrt{2 + \sqrt{3}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}})