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 | 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: $$E= \begin{aligned}[t] & ( c_1c_2-s_1s_2-c_1)(s_1c_2+c_1s_2+s_2) \\ &-(c_1c_2-s_1s_2-c_2)(s_1c_2+c_1s_2+s_1), \end{aligned} $$ with $c_i=\cos(t_i)$, $s_i=\sin(t_i)$, for $i=1,2$. Since $$c_1c_2-s_1s_2 = \cos(t_1+t_2)$$ and $$s_1c_2+c_1s_2=\sin(t_1+t_2),$$ we get $$E= \begin{aligned}[t] & \Bigl( \cos(t_1+t_2)-\cos(t_1)\Bigr) \Bigl( \sin(t_1+t_2)+\sin(t_2)\Bigr) \\ &-\Bigl( \cos(t_1+t_2)-\cos(t_2)\Bigr) \Bigl( \sin(t_1+t_2)+\sin(t_1)\Bigr). \end{aligned} $$ Once again, using well know trigonometric formulae, we obtain $$E= \begin{aligned}[t] & \Bigl(-2\sin\frac{2t_1+t_2}{2}\sin\frac{t_2}{2}\Bigr) \Bigl(2\sin\frac{t_1+2t_2}{2}\cos\frac{t_1}{2}\Bigr) \\ &-\Bigl(-2\sin\frac{t_1+2t_2}{2}\sin\frac{t_1}{2}\Bigr) \Bigl(2\sin\frac{2t_1+t_2}{2}\cos\frac{t_2}{2}\Bigr). \end{aligned} $$ Hence $$E= 4 \sin\frac{2t_1+t_2}{2} \sin\frac{t_1+2t_2}{2} \Bigl(\sin\frac{t_1}{2}\cos\frac{t_2}{2} -\cos\frac{t_1}{2}\sin\frac{t_2}{2}\Bigr),$$ that is, $$E= 4 \sin\frac{2t_1+t_2}{2}\ \sin\frac{t_1+2t_2}{2} \ \sin\frac{t_1-t_2}{2}.$$ Therefore, $E=0$ if and only if, for some $k\in\mathbb{Z}$, $2t_1+t_2=2k\pi$ or $t_1+2t_2=2k\pi$ or $t_1-t_2=2k\pi$. It is now quite obvious how to get $t_1$ in terms of $t_2$.
| | 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: $$E= \begin{aligned}[t] & ( c_1c_2-s_1s_2-c_1)(s_1c_2+c_1s_2+s_2) \\ &-(c_1c_2-s_1s_2-c_2)(s_1c_2+c_1s_2+s_1), \end{aligned} $$ with $c_i=\cos(t_i)$, $s_i=\sin(t_i)$, for $i=1,2$. Since $$c_1c_2-s_1s_2 = \cos(t_1+t_2)$$ and $$s_1c_2+c_1s_2=\sin(t_1+t_2),$$ we get $$E= \begin{aligned}[t] & \Bigl( \cos(t_1+t_2)-\cos(t_1)\Bigr) \Bigl( \sin(t_1+t_2)+\sin(t_2)\Bigr) \\ &-\Bigl( \cos(t_1+t_2)-\cos(t_2)\Bigr) \Bigl( \sin(t_1+t_2)+\sin(t_1)\Bigr). \end{aligned} $$ Once again, using well know trigonometric formulae, we obtain $$E= \begin{aligned}[t] & \Bigl(-2\sin\frac{2t_1+t_2}{2}\sin\frac{t_2}{2}\Bigr) \Bigl(2\sin\frac{t_1+2t_2}{2}\cos\frac{t_1}{2}\Bigr) \\ &-\Bigl(-2\sin\frac{t_1+2t_2}{2}\sin\frac{t_1}{2}\Bigr) \Bigl(2\sin\frac{2t_1+t_2}{2}\cos\frac{t_2}{2}\Bigr). \end{aligned} $$ Hence $$E= 4 \sin\frac{2t_1+t_2}{2} \sin\frac{t_1+2t_2}{2} \Bigl(\sin\frac{t_1}{2}\cos\frac{t_2}{2} -\cos\frac{t_1}{2}\sin\frac{t_2}{2}\Bigr),$$ that is, $$E= 4 \sin\frac{2t_1+t_2}{2}\ \sin\frac{t_1+2t_2}{2} \ \sin\frac{t_1-t_2}{2}.$$ Therefore, $E=0$ if and only if, for some $k\in\mathbb{Z}$, $2t_1+t_2=2k\pi$ or $t_1+2t_2=2k\pi$ or $t_1-t_2=2k\pi$. It is now quite obvious how to get $t_1$ in terms of $t_2$.
