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I was trying to think of a set of five trig functions in which the first trig function is multiplied by some constant a, the second by a different constant b, the third by a different constant c, the fourth by a different constant d, and the fifth by a different constant f, such that if $a^2+b^2+c^2+d^2+f^2=g^2$, the trig functions when multiplied by their constants, squared, and then added together also produce $g^2$. The trig functions I thought might be solutions to this don't work and I haven't been able to find any that work.

What five trig functions would be solutions to this problem.

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  • $\begingroup$ To be a little more clear and concise, you want five trig functions (say, $p(x)$, $q(x)$, $r(x)$, $s(x)$, $t(x)$) and five constants ($a$, $b$, $c$, $d$, $f$) such that $$a^2 + b^2 + c^2 + d^2 + f^2 \equiv (a\cdot p(x))^2 + (b \cdot q(x))^2 + (c \cdot r(x))^2+(d \cdot s(x))^2 + (f \cdot t(x))^2$$ Correct? Also: What makes you think that this problem has a solution? $\endgroup$ – Blue Nov 4 '17 at 20:43
  • $\begingroup$ Yes that is correct. I think the problem has a solution because there can be circles in five dimensions and if I'm not mistaken this would be the same equation as the equation for the square of the radius of a circle in 5d with $a*p(x)$ giving the first component for the coordinates of the circle, $b*q(x)$ giving the second component for the coordinates of the circle, $c*r(x)$ giving the third component for the coordinates of the circle, $d*s(x)$ giving the fourth component for the coordinates of the circle, and $f*t(x)$ giving the fifth component for the coordinates of the circle. $\endgroup$ – Anders Gustafson Nov 4 '17 at 20:59
  • $\begingroup$ The coords of a circle in $5$d may be more complicated than you suspect. Let $U:=(u_1,u_2,u_3,u_4,u_5)$, $V:=(v_1,v_2,v_3,v_4,v_5)$, $O:=(0,0,0,0,0)$. Take $|\overline{OU}|=|\overline{OV}|=r$, so that $$u_1^2+u_2^2+\cdots=v_1^2+v_2^2+\cdots=r^2$$ Further take $\overline{OU}\perp\overline{OV}$, so that $$u_1v_1+u_2v_2+\cdots = 0$$ The circle about $O$, through $U$ and $V$, has coords $U \cos t+ V \sin t$; each component looks like $u_i \cos t + v_i \sin t$, not simply "a constant, times a trig function". The final sum-of-squares is $$r^2=(u_1\cos t+v_1\sin t)^2+(u_1\cos t+v_1\sin t)^2+\cdots$$ $\endgroup$ – Blue Nov 4 '17 at 21:39
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I think what you're really looking for is the hyperspherical coordinate system in 5 dimensions.

$$\begin{eqnarray}a & = & g\cos(\phi_1) \\ b & = & g\sin(\phi_1)\cos(\phi_2) \\ c & = & g\sin(\phi_1)\sin(\phi_2)\cos(\phi_3) \\ d & = & g\sin(\phi_1)\sin(\phi_2)\sin(\phi_3)\cos(\phi_4) \\ f & = & g\sin(\phi_1)\sin(\phi_2)\sin(\phi_3)\sin(\phi_4) \\ \end{eqnarray}$$

with $\forall i \in {1,2,3}:\phi_i \in [0,\pi]$ and $\phi_4 \in [0,2\pi]$. Then, $a^2+b^2+c^2+d^2+f^2=g^2$.

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