Help me solve a trigonometric equation

Question

I am doing some work in RF circuit design. I need to solve an equation for my design:

$$\frac 1{\cos(t_1)}+\frac 1{\sin(t_1)} =\frac 1{\cos(t_2)}+\frac 1{\sin(t_2)}$$

(I created a nicely typed image of this equation but setting don't allow me to upload unless I get at least 10 !]

Kindly help me find a general solution of the above equation (i.e. to find relation between $t_2$ and $t_1$). I may solve easier equations (like $\,\cos t=0.5$ !), but seems complex to me. I observe that at $t_1=t_2=\dfrac {\pi}4$, this equation holds true. But I am unable to proceed further.

Thanks. Mohammad A Maktoomi.

Answer 1

Put $\sin\bigl(t+{\pi\over4}\bigr)=:u$. Then $${1\over\cos t}+{1\over\sin t}={\sqrt{2}\sin\bigl(t+{\pi\over4}\bigr)\over{1\over2}\sin(2t)}=-{2\sqrt{2} u\over\cos\bigl(2(t+{\pi\over4})\bigr)}=-2\sqrt{2}{u\over 1-2u^2}\ .$$ Disregarding the cases $u=\pm{1\over\sqrt{2}}$ we therefore have to consider the equation $${u_1\over 1-2u_1^2}={u_2\over 1-2u_2^2}\ ,$$ or $$(u_1-u_2)(1+2u_1u_2)=0\ .$$ There is the obvious solution $u_1=u_2$, which leads to (i) $t_1=t_2$ or (ii) $t_1+t_2={\pi\over2}$.

The solutions to $1+2u_1u_2=0$ can be parametrized by $$u_1=\pm {1\over\sqrt{2}} e^\tau\ ,\quad u_2=\mp {1\over\sqrt{2}} e^{-\tau}\qquad(-\log\sqrt{2}\leq \tau\leq\log\sqrt{2})\tag{1}$$ (note that $u_1$ and $u_2$ are sines). Each pair $(u_1,u_2)$ produced by $(1)$ will lead to two pairs $(t_1,t_2)$ solving the original equation; see the cases (i) and (ii) above.

Answer 2

Let's rewrite your equation : $$\frac {\sin(t_1)+\cos(t_1)}{\sin(t_1)\cos(t_1)} =\frac {\sin(t_2)+\cos(t_2)}{\sin(t_2)\cos(t_2)}$$ that you may multiply by $\frac 12\sin\left(\frac{\pi}4\right)$ and rewrite as : $$\frac {\sin\left(t_1+\frac {\pi}4\right)}{\sin(2\;t_1)} =\frac {\sin\left(t_2+\frac {\pi}4\right)}{\sin(2\;t_2)}$$ We could 'linearize' this to obtain : $$\sin\left(t_1+\frac {\pi}4\right)\sin(2\;t_2)=\sin\left(t_2+\frac {\pi}4\right)\sin(2\;t_1)$$

But it will be more convenient to set $\;x_1:=t_1+\frac {\pi}4,\ x_2:=t_2+\frac {\pi}4\;$ and get (since $\sin\bigl(2x-\frac {\pi}2\bigr)=-\cos(2x)$ and reverting the fractions) : $$\frac{\cos(2\;x_2)}{\sin(x_2)}=\frac{\cos(2\;x_1)}{\sin(x_1)}$$

and rewrite :

$$\frac{\cos(2\;x)}{\sin(x)}=\frac {1-2\sin(x)^2}{\sin(x)}=\frac 1{\sin(x)}-2\;\sin(x)$$ Since the solutions of $\frac 1s-2s=r$ are obtained by resolving $2s^2+rs-1=0$ with solutions $\;s=\frac{-r\pm\sqrt{8+r^2}}4$ we may deduce, for $s=\sin(x_2)$ and $r=\frac{\cos(2\;x_1)}{\sin(x_1)}$, that :

\begin{align} \sin(x_2)&=\frac{-\cos(2\;x_1)\pm\sqrt{8\;\sin(x_1)^2+\cos(2\;x_1)^2}}{4\;\sin(x_1)}\\ &=\frac{-\cos(2\;x_1)\pm\left(2-\cos(2\;x_1)\right)}{4\;\sin(x_1)}\\ \end{align} With the two different solutions : \begin{align} \sin(x_2)&=\frac{1-\cos(2\;x_1)}{2\;\sin(x_1)}=\sin(x_1)\\ \sin(x_2)&=-\frac{1}{2\;\sin(x_1)}\\ \end{align}

From this we deduce the four different solutions (modulo $2\pi$) :

\begin{align} x_2&=x_1\\ x_2&=-\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\ x_2&=\pi-x_1\\ x_2&=\pi+\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\ \end{align}

The substitution $t_1=x_1-\frac {\pi}4$ and $t_2=x_2-\frac {\pi}4$ should give you the wished solutions.