I have a couple of trigonometric elimination questions for you...

$1) a \sin \theta = b \sin (2 \theta)$; $c \cos \theta = d \cos (2 \theta)$ *NOTE: the first equation was incorrect...I had $\cos 2 \theta$ and it should be $\sin 2 \theta$.

I was able to get the answer to $2(abc - a^2 d + 2b^2 d)=0$, but the actual answer is $(c^2 - d^2)(abc - a^2 d + 2b^2 d)=0$. Where does the $(c^2 -d^2)$ come from?

UPDATE: I think I know $(c^2-d^2)$ comes from. If $\theta$ =0, then cos $\theta$ = 1 and in turn cos 2 $\theta$ = 1; plugging these values into the first equation gives us $c = d$. (I already did the second condition where $\frac {a} {2b}$)) If we square the first condition and subtract, that gives us $(c^2-d^2)$; since this also equals $2(abc - a^2 d + 2b^2 d)=0$, we can multiply both sides by $(c^2-d^2)$ to get the final result.

UPDATE #2: Actually, the more I thought about it, this is correct.

$2) \displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$

I'm not sure where to start...I've tried squaring both equations, multiplying, etc. (The answer for this is $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)

UPDATE: I was able to get the following for x and y (thanks to the suggestion!)

$\displaystyle x= 2\cos 2 \theta - \cos 4 \theta$

$\displaystyle y = 2\sin 2 \theta + \sin 4 \theta$

UPDATE 2/16/14: BREAKTHROUGH! Never mind...that answer bombed out.

UPDATE 2/17/14: Not giving up and using the advice below, I finally got $\displaystyle c = \frac {(x-1)^2 + (y^2-4)}{12}$...I'm going to plug this in to the equation for x and see how it goes. If all works well, I should finally get the answer, save all the cleanup work. Never mind.

UPDATE 4/8/17: I reposted question #2 to get a fresh perspective...see Trigonometric elimination (reprise from 2014). If it's a duplicate, my apologies!

Thanks for your help!

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    $\begingroup$ Congratulations on finding your solution! Please write it up in full as an answer here and accept it, so that your question doesn't remain in the Unanswered queue. $\endgroup$ – Blue Feb 12 '14 at 1:15

Here's a brute force approach to #2. For simplicity, I'll write $s$ and $c$ for $\sin\theta$ and $\cos\theta$.

  • Expand $\cos 3\theta$ and $\sin 3\theta$ $$\begin{align} x c + y s &= c \; ( 4 c^2 - 3 ) \\ x s - y c &= s \; ( 4 c^2 - 1 ) \end{align}$$

  • Isolate $s$ terms, and square, so that we can re-write in terms of $c$. $$\begin{align} (y s)^2 &= c^2 ( 4 c^2 - 3 - x )^2 \quad \to \quad y^2 ( 1 - c^2 ) - c^2 ( 4 c^2 - 3 - x )^2 = 0\\ (y c)^2 &= s^2 ( 4 c^2 - 1 - x )^2 \quad \to \quad y^2 c^2 - ( 1 - c^2 )( 4 c^2 - 1 - x )^2 = 0 \end{align}$$

  • Invoke one of my favorite tools ---the "method of resultants" (which I describe a bit in this answer)--- to eliminate $c$. Mathematica's (and/or WolframAlpha's) Resultant[] function makes it easy, yielding this polynomial equation: $$\begin{align}(\;x^4 - 4 y^4 + 3 x^2 y^2 + 8 x^3 - 18 x y^2 + 18 x^2 + 27 y^2 - 27 \;) & \\ \cdot\;(\;x^4 + y^4 + 2 x^2 y^2 + 8 x^3 - 24 x y^2 + 18 x^2 + 18 y^2 - 27 \;) &= 0\end{align}$$

  • The second factor corresponds to the answer you expect. Presumably, the first factor is extraneous in the current context (as is often the case with resultant results), but it's not obvious to me why that is.

The same process works for #1.

  • $\begingroup$ I tip my hat to you, Blue. This was a problem from Durell & Robson's Advanced Trigonometry and yours truly has been attempting to solve this for the last two months. I'll try it with both 1 and 2 and let you know. Thanks! $\endgroup$ – bjcolby15 Feb 18 '14 at 2:23
  • $\begingroup$ If the exercise is in a trig book (even an advanced one), chances are that you aren't expected to use stuff like resultants. (These days, I personally tend to hit most problems with the Resultant[] hammer first, and then look for more-elegant paths to the solution.) $\endgroup$ – Blue Feb 18 '14 at 2:33
  • $\begingroup$ For the record, I graduated with a BS in math 20 years ago! We were never taught resultants, so this is a new thing for me. You always learn something new! $\endgroup$ – bjcolby15 Feb 19 '14 at 2:21


$1)$ For the original version:

Divide to find $\tan\theta$

and use $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$

either $(i)$ in square one of the given equation

or $(ii)$ in $$\left(\frac{b\cos2\theta}b\right)^2+\left(\frac{d\cos2\theta}c\right)^2=\sin^2\theta+\cos^2\theta=\cdots$$

For the edited version

From the first relation, $\displaystyle\sin\theta(a-2b\cos\theta)=0$

If $\displaystyle\sin\theta=0,\cos\theta=\pm1,\cos2\theta=2\cos^2\theta-1=1$

Else $\displaystyle\cos\theta=\frac a{2b}$

Put the values of $\displaystyle\cos\theta,\cos2\theta$ in the second relation, $\displaystyle c\cos\theta=d\cos2\theta$


Solving for $x,y$

$\displaystyle x=2\cos2\theta-\cos4\theta=2c-(2c^2-1)\iff 2c^2=2c-1-x\ \ \ \ (A)$ where $c=\cos2\theta$

$\displaystyle y=2\sin2\theta-\sin4\theta=2\sin2\theta(1-\cos2\theta)$

$\displaystyle\implies y^2=4(1-c^2)(1-c)^2\ \ \ \ (B)$

Put the value of $2c^2$ from $(A)$ in $(B)$ to find $c$ in terms of $y^2$

Then, put this value of $c$ in $(A)$


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