Answer: sin2y =240/289 where 0<y <π/2
Step-by-step explanation:
Given sin(x) = 1/3 and sec(y) = 17/15 , where x and y lie between 0 and π/2.
We now that
[tex]cosy = \frac{1}{secy} =\frac{15}{17} \\and\\\text{using identity } sin^2y+cos^2y=1\\\text{we have}\\siny=\sqrt{1-cos^2y}=\sqrt{1-(\frac{15}{17} )^2}=\sqrt{1-\frac{225}{289} }=\sqrt{\frac{64}{289} }=\frac{8}{17} \\\Rightarrow\ siny=\frac{8}{17} \\\text{Now we are using identity sin2A=2sinAcosA we have}\\sin2y=2\cdot\ siny\cdot\ cosy=2\cdot\ \frac{8}{17} \cdot\ \frac{15}{17} =\frac{240}{289}[/tex]
