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Arada [10]
3 years ago
12

Find tan2θ if θ terminates in Quadrant IV and cosθ = 3/5.

Mathematics
2 answers:
natali 33 [55]3 years ago
4 0

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\cos( \alpha )  =  \frac{3}{5}   \\

{sin}^{2}( \alpha )  = 1 -  {cos}^{2}  ( \alpha )

{sin}^{2}( \alpha )  = 1 -  ({ \frac{3}{5} })^{2} \\

{sin}^{2}( \alpha )   = 1 -  \frac{9}{25}  \\

{sin}^{2}( \alpha )   =  \frac{25}{25}  -  \frac{9}{25}  \\

{sin}^{2}( \alpha ) =  \frac{16}{25}   \\

sin( \alpha )= ± \sqrt{ \frac{16}{25} }  \\

In Quadrant IV , sin is negative .

Thus ;

sin( \alpha ) =  -  \sqrt{ \frac{16}{25} }  \\

\sin( \alpha )  =  -  \frac{4}{5}  \\

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\tan(2 \alpha )  =  \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }  \\

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\sin(2 \alpha ) = 2. \sin( \alpha ).  \cos( \alpha )

\sin(2 \alpha )  = 2 \times ( -  \frac{4}{5} ) \times ( \frac{3}{5} ) \\

\sin(2 \alpha )  =  -  \frac{24}{25}  \\

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\cos(2 \alpha ) =  {cos}^{2}( \alpha ) -  {sin}^{2}( \alpha )

\cos(2 \alpha ) =  ({ \frac{3}{5} })^{2} -  ({ -  \frac{4}{5} })^{2}   \\

\cos(2 \alpha )  =  \frac{9}{25}  -  \frac{16}{25}  \\

\cos(2 \alpha )  =  -  \frac{7}{25}  \\

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\tan(2 \alpha )  =  \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }  \\

\tan(2 \alpha )  =  \frac{ -  \frac{24}{25} }{ -  \frac{7}{25} }  \\

\tan(2 \alpha ) =  -  \frac{24}{25}    \div   -  \frac{7}{25}  \\

\tan(2 \alpha )  =  -  \frac{24}{25}  \times   -  \frac{25}{7}  \\

\tan(2 \alpha )  =  \frac{24}{7}  \\

Thus the correct answer is (( C )) .

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Doss [256]3 years ago
3 0

Answer:

we have

cos θ = 3/5.

sin²θ=1-cos²θ

sin²θ=1-9/25

sin²θ=16/25

sinθ=±4/5

since it lies in IV qyadrant so

sinθ=-4/5

we have

tan2θ=sin2θ/cos2θ=2sinθcosθ/[cos²θ-sin²θ)

=(2×-4/5×3/5)/(9/25-16/25)=24/7

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