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

Question

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.

Answer 1

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

$\dfrac{\sin\theta}{(\sin\alpha)^{1/3}}=\dfrac{\cos\theta}{(\cos\alpha)^{1/3}}=\pm\dfrac1{\sqrt{(\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}}}$

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

$$\lambda(\cos^2\theta-\sin^2\theta)=\cos\theta\cos\alpha-\sin\theta\sin\alpha$$

$$\implies\lambda\dfrac{(\cos\alpha)^{2/3}-(\sin\alpha)^{2/3}}{(\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}}=\pm\dfrac{(\cos\alpha)^{1/3+1}-(\sin\alpha)^{1/3+1}}{\sqrt{(\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}}}$$

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

$$\lambda=\pm\left((\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}\right)^{1+1-1/2}$$

$$\implies\lambda^{2/3}=(\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}$$

Answer 2

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

$$\tan^2\alpha+\frac{1}{\tan^2\alpha}=\tan^6\theta+\frac{1}{\tan^6\theta}=\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)^3-3\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)$$

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

Answer 3

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 $|\lambda|\geq 1$.

Answer 4

Result

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

$$\sin^2(\theta+\alpha)=\frac{\lambda^2-1}{3}\tag{2}$$

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 $(1)$.

Answer 5

^{I'm late here and @lab bhattacharjee has already mentioned the beautiful answer to this question. I am only writing this to add another way to reach that answer.}

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Expanding $\cos(\theta+\alpha)$ and $\sin(\theta+\alpha)$, our equations can be written as, $$\lambda\cos2\theta=\cos\theta\cos\alpha-\sin\theta\sin\alpha$$ $$\lambda\sin2\theta=2\sin\theta\cos\alpha+2\cos\theta\sin\alpha$$

Now we solve for $\cos\alpha$, $$\lambda(2\cos\theta\cos2\theta+\sin\theta\sin2\theta)=2\cos\alpha(\cos^2\theta+\sin^2\theta)$$ $$\lambda(\cos\theta\cos2\theta+\cos\theta)=2\cos\alpha$$ $$\lambda\cos\theta(\underbrace{\cos2\theta+1}_{2\cos^2\theta})=2\cos\alpha$$ $$\lambda\cos^3\theta=\cos\alpha\tag1$$

Similarly, solving for $\sin\alpha$, we get, $$\lambda\sin^3\theta=\sin\alpha\tag2$$

Finally, using the famous identity $\cos^2\theta+\sin^2\theta=1$,

$$\lambda^{2/3}=(\cos\alpha)^{2/3}+(\sin\alpha)^{2/3}$$

$\blacksquare$