How do you solve $x=h+acos(t)cos(s)-bsin(t)sin(s)$ for $t$ ?

Question

Show steps.

1495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-09T01:09:16

See explicit t = g ( x, s )#, in the explanation.

Using c = cos s and d = sin s,

$( x - h )^2 = c^2a^2( 1 - d^2 ) + ( 1 - c^2 ) b^2 d^2$

$- 2cdab sqrt(( 1 - c^2 )( 1 - d^2))$ . Reorganizing and squaring,

$(( x - h )^2 - c^2a^2( 1 - d^2 ) - ( 1 - c^2 ) b^2 d^2)^2$

$= 4(dab)^2(1-d^2) c^2 ((1-c^2)$

Befitting ( $abs c <= 1$ ) solutions ${ cos alpha }$ of this

biquadratic in c, lead to

$c = cos t = cos alpha$

$rArr$ , piecewise,

$t = 2kpi +-alpha$ rad.

Example: a = b = 1 and h = 0.

$x = cos t cos s - sin t sin s$ and

(x^2 -c^2 - d^2 + 2c^2 d^2 )^2#

$= 4 c^2 (1-c^2) d^2 (1-d^2)$

Solve this quadratic in $c^2$ for c = cos t.

I think, I have paved the way.