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answered 5 years ago

Juanjo gravatar image

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=(c1c2s1s2c1)(s1c2+c1s2+s2)(c1c2s1s2c2)(s1c2+c1s2+s1), with ci=cos(ti), si=sin(ti), for i=1,2. Since c1c2s1s2=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(sint12cost22cost12sint22), that is, E=4sin2t1+t22 sint1+2t22 sint1t22. Therefore, E=0 if and only if, for some kZ, 2t1+t2=2kπ or t1+2t2=2kπ or t1t2=2kπ. It is now quite obvious how to get t1 in terms of t2.

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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=(c1c2s1s2c1)(s1c2+c1s2+s2)(c1c2s1s2c2)(s1c2+c1s2+s1), with ci=cos(ti), si=sin(ti), for i=1,2. Since c1c2s1s2=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(sint12cost22cost12sint22), that is, E=4sin2t1+t22 sint1+2t22 sint1t22. Therefore, E=0 if and only if, for some kZ, 2t1+t2=2kπ or t1+2t2=2kπ or t1t2=2kπ. It is now quite obvious how to get t1 in terms of t2.