Given $sin θ = \frac{4}{5}$ $sin theta=4/5$ , $θ$ $theta$ lies in Quadrant 2, find $\frac{cos θ}{2}$ $cos theta/2$ ?

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Given $sin θ = \frac{4}{5}$ $sin theta=4/5$ , $θ$ $theta$ lies in Quadrant 2, find $\frac{cos θ}{2}$ $cos theta/2$ ?

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Answer 1 · 2018-03-14T00:15:50

$\frac{cos θ}{2} = - \frac{3}{10}$ $costheta/2= -3/10$

Explanation:

It lies in the second quadrant
So here is what I know:
Sin(y-value) is positive
Cos(x-value) is negative

I also know that $sin θ$ $sintheta$ is Opposite/Hypotenuse
Therefore: $4$ $4$ is the length of the opposite leg and the length of the hypotenuse is $5$ $5$

We can either use Pythagorean theorem to find the length of the adjacent leg or we can apply a Pythagorean triple:
Pythagorean Theorem: $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$
Manipulated to: $b = \pm \sqrt{c^{2} - a^{2}}$ $b=+-sqrt(c^2-a^2)$
$b = \pm \sqrt{{(5)}^{2} - {(4)}^{2}}$ $b=+-sqrt((5)^2-(4)^2)$
$b = \pm \sqrt{25 - 16}$ $b=+-sqrt(25-16)$
$b = \pm \sqrt{9}$ $b=+-sqrt(9)$
$b = \pm 3$ $b=+-3$
Since 4 is supposed to be the Cos value, we will use $- 3$ $-3$

$cos θ = - \frac{3}{5}$ $costheta=-3/5$

Therefore: $\frac{cos θ}{2} = \frac{- \frac{3}{5}}{2}$ $costheta/2= (-3/5)/2$

$\frac{cos θ}{2} = - \frac{3}{10}$ $costheta/2= -3/10$

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Given $sin θ = \frac{4}{5}$ $sin theta=4/5$ , $θ$ $theta$ lies in Quadrant 2, find $\frac{cos θ}{2}$ $cos theta/2$ ?

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Given $sin θ = \frac{4}{5}$ $sin theta=4/5$ , $θ$ $theta$ lies in Quadrant 2, find $\frac{cos θ}{2}$ $cos theta/2$ ?