Question

Given cosθ= -gn="absmiddle" class="latex-formula"> and tanθ<0, find sin2θ

Answer 1

$\sin {}^{2} (x) + {cos}^{2} (x) = 1$

${sin}^{2} (x) = 1 - {cos}^{2} (x)$

${sin}^{2} (x) = 1 - ({ \frac{ - 7}{9} })^{2}$

${sin}^{2} (x) = 1 - \frac{49}{81}$

${sin}^{2} (x) = \frac{81}{81} - \frac{49}{81}$

${sin}^{2} (x) = \frac{32}{81}$

$\sin(x) = ± \sqrt{ \frac{32}{81} }$

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$\tan(x) < 0 \: \: \: and \: \cos(x) < 0$

Thus :

$\sin(x) > 0$

So we have :

$\sin(x) = + \sqrt{ \frac{32}{81} }$

$\sin(x) = \frac{ \sqrt{16 \times 2} }{ \sqrt{9 \times 9} }$

$\sin(x) = \frac{4 \sqrt{2} }{9}$

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$\sin(2x) = 2 \times \sin(x) \times \cos(x)$

$\sin(2x) = 2 \times \frac{4 \sqrt{2} }{9} \times ( - \frac{7}{9} )$

$\sin(2x) = - \frac{56 \sqrt{2} }{81}$