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

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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.