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Answer 1 · 2017-07-10T15:13:52

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$tan x = - \frac{9}{40}$ $tan x = -9/40$
${csc}^{2} x = 1 + {cot}^{2} x$ $csc^2x = 1+cot^2x$

${csc}^{2} x = 1 + {(- \frac{40}{9})}^{2}$ $csc^2x = 1+(-40/9)^2$

${csc}^{2} x = \frac{1681}{81}$ $csc^2x = 1681/81$

$csc x = \pm \frac{41}{9}$ $csc x = \pm 41/9$

Because it should be in 4th quadrant we have

$csc x = - \frac{41}{9}$ $csc x = -41/9$

$sin x = - \frac{9}{41}$ $sinx = -9/41$

$2 sin (\frac{x}{2}) cos (\frac{x}{2}) = - \frac{9}{41}$ $2sin(x/2)cos(x/2) = -9/41$

$sin (\frac{x}{2}) \sqrt{1 - {sin}^{2} (\frac{x}{2})} = - \frac{9}{82}$ $sin(x/2)sqrt(1-sin^2(x/2)) = -9/82$

Squaring both sides we get

${sin}^{2} (\frac{x}{2}) (1 - {sin}^{2} (\frac{x}{2})) = {(- \frac{9}{82})}^{2}$ $sin^2(x/2)(1-sin^2(x/2)) = (-9/82)^2$

Let $t = {sin}^{2} (\frac{x}{2})$ $t = sin^2(x/2)$

$t (1 - t) - {(\frac{9}{82})}^{2} = 0$ $t(1-t)-(9/82)^2=0$

$t^{2} - t + {(\frac{9}{82})}^{2} = 0$ $t^2-t+(9/82)^2=0$

$t = \frac{1 \pm \sqrt{1 - 4 {(\frac{9}{82})}^{2}}}{2}$ $t = (1\pm sqrt(1-4(9/82)^2))/2$

$t = \frac{1 \pm (\frac{40}{41})}{2}$ $t = (1\pm(40/41))/2$

$t = \frac{81}{82}, \frac{1}{82}$ $t = 81/82,1/82$

$sin (\frac{x}{2}) = - \frac{9}{\sqrt{82}}, - \frac{1}{\sqrt{82}}$ $sin(x/2) = -9/sqrt(82),-1/sqrt(82)$