Answer:
[tex]sin(\theta) = \frac{\sqrt{17} }{7}[/tex]
Step-by-step explanation:
We know that:
sin(x)^2 + cos(x)^2 = 1
And we know that:
[tex]cos(\theta) = \frac{4\sqrt{2}}{7} }[/tex]
We want to find the value of the sine function evaluated in theta.
If we replace that in the first equation, we get:
[tex]sin(\theta)^2 + cos(\theta)^2 = 1[/tex]
[tex]sin(\theta)^2 + (\frac{4*\sqrt{2} }{7}) ^2 = 1[/tex]
[tex]sin(\theta)^2 + (\frac{4^2*\sqrt{2}^2 }{7^2}) = 1[/tex]
[tex]sin(\theta)^2 + (\frac{16*2 }{49}) = 1[/tex]
Now we can just isolate the sine part of that equation, so we get:
[tex]sin(\theta)^2 = - (\frac{16*2 }{49}) + 1 = \frac{-32}{49} + \frac{49}{49} = \frac{-32 + 49}{49} = \frac{17}{49}[/tex]
[tex]sin(\theta) = \sqrt{\frac{17}{49} } = \frac{\sqrt{17} }{\sqrt{49} } = \frac{\sqrt{17} }{7}[/tex]
(We can't simplify the fraction anymore)
