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Natali [406]
2 years ago
5

Given cosθ= -

gn="absmiddle" class="latex-formula"> and tanθ<0, find sin2θ
Mathematics
1 answer:
Dafna1 [17]2 years ago
5 0

\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} }  \\

_____________________________________________

\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}  \\

_____________________________________________

\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}  \\

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