Given cosθ= -(7)/(9) and tanθ<0, find sin2θ

Question

Given cosθ= -
`(7)/(9)` 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 - ({ ( - 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) \\`