Question #5b8a3

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Answer 1 · 2016-12-11T04:48:35

$(r, θ) = (3.1623, - {18.435}^{o})$ $(r, theta)=(3.1623, -18.435^o)$

The conversion formula $(x, y) = r (cos θ, sin θ)$ $(x, y)=r(cos theta, sin theta)$ .

Here, $x = r cos θ = - 1 and y = r sin θ = 3$ $x = r cos theta=-1 and y = r sin theta = 3$

So,

$r = \sqrt{x^{2} + y^{2}} = \sqrt{10} = 3.1623$ $r =sqrt(x^2+y^2)=sqrt10=3.1623$ , nearly,

$cos θ = \frac{x}{r} = - \frac{1}{\sqrt{10}} and sin θ = \frac{3}{\sqrt{10}}$ $cos theta = x/r=-1/sqrt10 and sin theta =3/sqrt10$ .

The prefixing signs $- +$ $- +$ reveal that $θ \in Q_{4}$ $theta inQ_4$ .

The calcular value ${sin}^{- \frac{1}{\sqrt{10}}} = - {18.435}^{o} \in Q_{4}$ $sin^(-1/sqrt10)=-18.435^o in Q_4$

So, $(r, θ) = (3.1623, - {18.435}^{o})$ $(r, theta)=(3.1623, -18.435^o)$ .

Note: Had you used the cosine value, the calculator display will be

+18.435. But cos(-theta)=cos(theta) and we can change the sign for

the shift to $Q_{4}$ $Q_4$ , wherein sine is negative. If you require

a positive equivalent of $θ$ $theta$ , for the same direction,

it is $(360 - 18.435) = 341.565$ $(360-18.435)=341.565$ , the coordinates become

$(3.1623, {341.565}^{o})$ $(3.1623, 341.565^o)$