![]() | 1 | initial version |
It seems that the opening post is homework, since the equation can be solve by hand. Anyway, let us see how.
Let E be the left-hand side of the equation given in the opening post. It can be rewritten as follows: E=(c1c2−s1s2−c1)(s1c2+c1s2+s2)−(c1c2−s1s2−c2)(s1c2+c1s2+s1), with ci=cos(ti), si=sin(ti), for i=1,2. Since c1c2−s1s2=cos(t1+t2) and s1c2+c1s2=sin(t1+t2), we get E=(cos(t1+t2)−cos(t1))(sin(t1+t2)+sin(t2))−(cos(t1+t2)−cos(t2))(sin(t1+t2)+sin(t1)). Once again, using well know trigonometric formulae, we obtain E=(−2sin2t1+t22sint22)(2sint1+2t22cost12)−(−2sint1+2t22sint12)(2sin2t1+t22cost22). Hence E=4sin2t1+t22sint1+2t22(sint12cost22−cost12sint22), that is, E=4sin2t1+t22 sint1+2t22 sint1−t22. Therefore, E=0 if and only if, for some k∈Z, 2t1+t2=2kπ or t1+2t2=2kπ or t1−t2=2kπ. It is now quite obvious how to get t1 in terms of t2.
![]() | 2 | No.2 Revision |
It seems that the opening post is homework, since the equation can be solve solved by hand. hand, if I am right. Anyway, let us see how.
Let E be the left-hand side of the equation given in the opening post. It can be rewritten as follows: E=(c1c2−s1s2−c1)(s1c2+c1s2+s2)−(c1c2−s1s2−c2)(s1c2+c1s2+s1), with ci=cos(ti), si=sin(ti), for i=1,2. Since c1c2−s1s2=cos(t1+t2) and s1c2+c1s2=sin(t1+t2), we get E=(cos(t1+t2)−cos(t1))(sin(t1+t2)+sin(t2))−(cos(t1+t2)−cos(t2))(sin(t1+t2)+sin(t1)). Once again, using well know trigonometric formulae, we obtain E=(−2sin2t1+t22sint22)(2sint1+2t22cost12)−(−2sint1+2t22sint12)(2sin2t1+t22cost22). Hence E=4sin2t1+t22sint1+2t22(sint12cost22−cost12sint22), that is, E=4sin2t1+t22 sint1+2t22 sint1−t22. Therefore, E=0 if and only if, for some k∈Z, 2t1+t2=2kπ or t1+2t2=2kπ or t1−t2=2kπ. It is now quite obvious how to get t1 in terms of t2.