Choose y∈R so that sin⁡y=uu2+v2 and cos⁡y=vu2+v2I know the identity sin2y+cos2y=1. Also, I notice that(uu2+v2)2+(vu2+v2)2=1without u,v vanishing simultaneously. But I am not sure whether ∃y∈R s.t. sin⁡y=uu2+v2 and cos⁡y=vu2+v2.Does anyone have an idea?

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Curtiello69r · Accepted Answer

The identity(uu2+v2)2+(vu2+v2)2=1means that the point (uu2+v2,vu2+v2) is on the unit circle x2+y2=1. By definition, there exists an angle y (actually, an infinity of angles) such that cos⁡(y)=uu2+v2 and sin⁡(y)=vu2+v2Assume for example that u&amp;gt;0 and v&amp;gt;0(*), so that y is in the first quadrant. Then, y=arcsin⁡(vu2+v2)+2kπ and x=arccos⁡(uu2+v2)+2kπ where k is an integer.(*) For other possible signs of u and v, we obtain similar formulas.

Dixie Klein · Accepted Answer

It is clear that vu2+v2∈[−1,1]. Therefore, since cos⁡0=1  and  cos⁡π=−1, the intermediate value theorem implies that there&#039;s a y∈[0,π] such that cos⁡y=vu2+v2. Thensin2y=1−cos2y=1−v2u2+v2=(uu2+v2)2and therefore sin⁡y=±uu2+v2. If you got the - sign, then replace y by -y and you&#039;re done.

Choose \(\displaystyle{y}\in{\mathbb{{{R}}}}\) so that \(\displaystyle{\sin{{y}}}={\frac{{{u}}}{{\sqrt{{{u}^{{2}}+{v}^{{2}}}}}}}\) and

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