Eliminate $\theta$ from $\lambda \cos2\theta=\cos(\theta + \alpha)$ and $\lambda\sin2\theta=2\sin(\theta + \alpha)$

My approach:

Dividing the RHS and LHS of both equations by $\lambda$, then squaring and adding them, we get, $$\frac{\cos^2(\theta+\alpha)}{\lambda^2}+\frac{4\sin^2(\theta+\alpha)}{\lambda^2}=\cos^22\theta + \sin^22\theta=1$$ $$\Rightarrow \sin^2(\theta+\alpha)=\frac{\lambda^2-1}{3}$$ I am unable to proceed.


We have

\begin{align*} \tan2\theta&=2\tan(\theta+\alpha)\\ \frac{2\tan\theta}{1-\tan^2\theta}&=\frac{2(\tan\theta+\tan\alpha)}{1-\tan\theta\tan\alpha}\\ \tan\theta-\tan^2\theta\tan\alpha&=\tan\theta(1-\tan^2\theta)+\tan\alpha(1-\tan^2\theta)\\ \tan^3\theta&=\tan\alpha \end{align*}

From $\lambda\sin2\theta=2\sin(\theta+\alpha)$,

\begin{align*} 2\lambda\sin\theta\cos\theta&=2(\sin\theta\cos\alpha+\cos\theta\sin\alpha)\\ \lambda&=\frac{\cos\alpha}{\cos\theta}+\frac{\sin\alpha}{\sin\theta}\\ &=\frac{\cos\alpha}{\cos\theta}\left(1+\frac{\tan\alpha}{\tan\theta}\right)\\ \lambda^2&=\frac{\sec^2\theta}{\sec^2\alpha}\left(1+\frac{\tan\alpha}{\tan\theta}\right)^2\\ &=\left(\frac{1+\tan^2\theta}{1+\tan^2\alpha}\right)(1+\tan^2\theta)^2\\ &=\frac{(1+\tan^2\theta)^3}{1+\tan^6\theta}\\ &=\frac{1+2\tan^2\theta+\tan^4\theta}{1-\tan^2\theta+\tan^4\theta}\\ (\lambda^2-1)(1+\tan^4\theta)&=(\lambda^2+2)\tan^2\theta\\ \tan^2\theta+\frac{1}{\tan^2\theta}&=\frac{\lambda^2+2}{\lambda^2-1} \end{align*}

Note that


Therefore, $\displaystyle \tan^2\alpha+\frac{1}{\tan^2\alpha}=\left(\frac{\lambda^2+2}{\lambda^2-1}\right)^3-3\left(\frac{\lambda^2+2}{\lambda^2-1}\right)$.


From $\tan^3\theta=\tan\alpha$


From $\lambda\cos2\theta=\cos(\theta+\alpha)$



Assuming $\cos\alpha\ne\sin\alpha,$



  • $\begingroup$ @CYAries, Could you please verify? $\endgroup$ – lab bhattacharjee May 4 at 16:16
  • $\begingroup$ This is beautiful. $\endgroup$ – CY Aries May 6 at 1:24
  • $\begingroup$ It fits the identity I found. But this solution is much better. $\endgroup$ – CY Aries May 6 at 1:35

Take the square root and then apply $\arcsin$ (odd function) to both sides of the equation. You will obtain

$$\theta = \pm \arcsin \sqrt{\dfrac{\lambda^2-1}{3}}$$ $$\theta = - \alpha \pm \arcsin \sqrt{\dfrac{\lambda^2-1}{3}}$$

Note, that by taking the square root of $\lambda^2-1$ we restrict the possible values of $\lambda$. We have $|\lambda|\geq 1$.

  • $\begingroup$ We have been asked to eliminate $\theta$ not to solve for it. This means you need to find a third equation relating all variables except $\theta$ $\endgroup$ – ami_ba May 1 at 17:00
  • 2
    $\begingroup$ After solving for $\theta$ we can remove it in the initial equations. $\endgroup$ – MachineLearner May 1 at 17:02
  • $\begingroup$ @MachineLearner, We have $$\sin^22\theta=\dfrac{4(\lambda^2-1)}{3\lambda^2}$$ then $\theta=?$ $\endgroup$ – lab bhattacharjee May 4 at 15:56


First we calculate $\theta$

$$\theta = \frac{1}{2} \arcsin\left(\frac{4}{3}(1-\frac{1}{\lambda^2})\right)\tag{1}$$

if $\lambda^2 >1$, and else no solution.

This can be combined with the equation derived in the OP


to solve for $\alpha$ so that both quantities are eliminated and the equations are completely solved in terms of $\lambda$.

Derivation of (1)

We have

$$ \cos(\theta+\alpha)=\lambda \cos(2 \theta)$$ $$ \sin(\theta+\alpha)=\frac{1}{2}\lambda \sin(2 \theta)$$

so that

$$1 = \lambda ^2 \cos(2 \theta)^2 + \frac{1}{4} \lambda ^2 \sin(2 \theta)^2$$

which eliminates $\alpha$.

Hence, observing $\cos(2 \theta)^2 + \sin(2 \theta)^2 = 1$, follows $(1)$.

  • $\begingroup$ We need to eliminate $\theta$ $\endgroup$ – ami_ba May 1 at 17:06
  • $\begingroup$ Thanks for the hint. I have now eliminated both $\theta$ and $\alpha$ and expressed them through $\lambda$, i.e. solved the system completely. $\endgroup$ – Dr. Wolfgang Hintze May 1 at 17:12
  • $\begingroup$ The elimination problem makes more sense if you drop the second equation. You then get an algebraic equation of 8th degree for $\sin(\theta)$. $\endgroup$ – Dr. Wolfgang Hintze May 1 at 17:24

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