Given
[tex]\cos x=\frac{2}{5},270\degree
Find[tex]\sin2x[/tex]
Explanation
As we know that
[tex]\sin2x=2\sin x\cos x[/tex]
and
[tex]\cos x=\frac{B}{H}[/tex]
H = 5 and B = 2
By pythagoras theorem ,
[tex]\begin{gathered} H^2-B^2=P^2 \\ 5^2-2^2=P^2 \\ 25-4=P^2 \\ P=\sqrt{21} \end{gathered}[/tex]
so ,
[tex]\sin x=\frac{P}{H}=\frac{\sqrt{21}}{5}[/tex]
as x lies in 4th quadrant , and we know sinx is negative in 4th quadrant.
so ,
[tex]\begin{gathered} \sin2x=2\sin x\cos x \\ \sin2x=2(-\frac{\sqrt{21}}{5})(\frac{2}{5}) \\ \\ \sin2x=-\frac{4\sqrt{21}}{25} \end{gathered}[/tex]
Final Answer
Hence , the correct option is 4.
