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I'm struggling to solve a simple set of two equations. I've tried rearranging the first one and substituting into the second but I can't seem to get an answer.

I want to solve for C and D.

$$ C\sin(2s)+D\cos(2s)=0 $$ $$ 2C\cos(2s)+2D\sin(2s)=1 $$

I've tried rearranging the top one for $C$ and substituting into the second, but I think I am missing a simplification to get an answer.

Any help would be great, Thanks!

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  • $\begingroup$ Square both equations and add them? $\endgroup$ – juper Mar 28 '18 at 15:29
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HINT. Note first that necessarily $C\ne D$. Note $x=\sin(2s)$ and $y=\cos(2s)$.

The system is equivalent to $$\frac{Cx}{\sqrt{C^2+D^2}}+\frac{Dy}{\sqrt{C^2+D^2}}=0$$ $$\dfrac{2Dx}{\sqrt{4(C^2+D^2)}}+\dfrac{2Cy}{\sqrt{4(C^2+D^2)}}=1$$

Put $\sin \theta_1=\dfrac{Dy}{\sqrt{C^2+D^2}}$ and $\sin \theta_2=\dfrac{Cy}{\sqrt{C^2+D^2}}$ so you have $$\cos\theta_1\sin 2s+\sin\theta_1\cos 2s=\sin(\theta_1+2s)=0$$ Or similarly $$\cos\theta_2\sin 2s+\sin\theta_2\cos 2s=\sin(\theta_2+2s)=1$$ so you have

$$\sin(\theta_1+2s)=0\iff\theta_1+2s=\arcsin (0)\\\sin(\theta_2+2s)=1\iff\theta_2+2s=\arcsin (1)$$

On the other hand, solving the given system, you have the relation $$D^2+C^2=4(C^2-D^2)^2$$

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From the first equation we get $$D=-C\tan(2s)$$ plugging this in the second one we obtain $$2C\cos(2s)-2C\tan(2s)\sin(2s)=1$$ solving this for $C$ we get $$C=\frac{1}{2\cos(2s)-2\tan(2s)\sin(2s)}$$ $$C=\frac{\cos(2s)}{2(\cos^2(2s)-\sin^2(2s))}$$ so $$D=-\frac{\sin(2s)}{2(\cos^2(2s)-\sin^2(2s))}$$

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  • $\begingroup$ My equations are part of a larger problem I'm working on where D can equal 0. Does this change the above? $\endgroup$ – juper Mar 28 '18 at 15:28
  • $\begingroup$ How do you solve $C$ and $D$ with the last eq? $\endgroup$ – John Glenn Mar 28 '18 at 17:01
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From $$C\sin(2s)+D\cos(2s)=0$$ $$2C\cos(2s)+2D\sin(2s)=1$$ You can try to divide both equations by $\cos 2s$, and you get: $$C\tan2s+D=0$$ $$2C+2D\tan2s=\sec2s$$ You get $D=-C\tan2s$ and substitute in the second: $$2C+2(-C\tan2s)\tan2s=\sec2s\Rightarrow C=\frac{\sec2s}{2-2\tan^22s}$$ Then you get $D$ as: $$D=-C\tan2s\Rightarrow D=-\frac{\sec2s}{2-2\tan^22s}\cdot\tan2s$$ $$D=\frac{\frac{\sin2s}{\sec^22s}}{2-2\tan^22s}=\frac{\frac{\sin2s}{1+\tan^22s}}{2-2\tan^22s}=\frac{\sin2s}{2-2\tan^42s}$$

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  • $\begingroup$ Dividing by $\cos x$ to get $\tan x$ sometimes helps avoid the need to square equations. (Not that it was needed here.) $\endgroup$ – John Glenn Mar 28 '18 at 17:25
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You can set $C = \lambda \cos(2 s)$ and $S =-\lambda \sin(2 s)$ since that would solve the first equation

$$ C \sin(2 s)+D \cos(2 s) = \lambda \cos(2 s) \sin(2 s) - \lambda \sin(2 s) \cos(2 s) = 0 $$

Use the second equation to find $\lambda$

$$ 2 C \cos(2 s) + 2 D \sin(2 s) = 2 \lambda \cos^2(2 s) - 2 \lambda \sin^2(2 s) = 1 $$

$$ \lambda = \frac{1}{2 \left( 2 \cos^2(2 s)-1\right)} = \frac{1}{2 \cos(4 s)} $$

The solution is

$$ \begin{aligned} C & = \frac{\cos(2 s)}{2 \cos(4 s)} \\ D & =- \frac{\sin(2 s)}{2 \cos(4 s)} \end{aligned} $$

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