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Given cosθ= -[tex]\frac{7}{9}[/tex] and tanθ<0, find sin2θ

Sagot :

[tex] \sin {}^{2} (x) + {cos}^{2} (x) = 1[/tex]

[tex] {sin}^{2} (x) = 1 - {cos}^{2} (x)[/tex]

[tex] {sin}^{2} (x) = 1 - ({ \frac{ - 7}{9} })^{2} \\ [/tex]

[tex] {sin}^{2} (x) = 1 - \frac{49}{81} \\ [/tex]

[tex] {sin}^{2} (x) = \frac{81}{81} - \frac{49}{81} \\ [/tex]

[tex] {sin}^{2} (x) = \frac{32}{81} \\ [/tex]

[tex] \sin(x) = ± \sqrt{ \frac{32}{81} } \\ [/tex]

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[tex] \tan(x) < 0 \: \: \: and \: \cos(x) < 0[/tex]

Thus :

[tex] \sin(x) > 0[/tex]

So we have :

[tex] \sin(x) = + \sqrt{ \frac{32}{81} } \\ [/tex]

[tex] \sin(x) = \frac{ \sqrt{16 \times 2} }{ \sqrt{9 \times 9} } \\ [/tex]

[tex] \sin(x) = \frac{4 \sqrt{2} }{9} \\ [/tex]

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[tex] \sin(2x) = 2 \times \sin(x) \times \cos(x) [/tex]

[tex] \sin(2x) = 2 \times \frac{4 \sqrt{2} }{9} \times ( - \frac{7}{9} ) \\ [/tex]

[tex] \sin(2x) = - \frac{56 \sqrt{2} }{81} \\ [/tex]