Question 1

Given cosθ= -
(7)/(9)

and tanθ<0, find sin2θ

Question 2

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

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

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

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

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

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

$\sin(x) = ± \sqrt{ (32)/(81) } \\$

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

Thus :

$\sin(x) > 0$

So we have :

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

$\sin(x) = ( √(16 * 2) )/( √(9 * 9) ) \\$

$\sin(x) = (4 √(2) )/(9) \\$

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

$\sin(2x) = 2 * (4 √(2) )/(9) * ( - (7)/(9) ) \\$

$\sin(2x) = - (56 √(2) )/(81) \\$

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