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Find the equation of each tangent to the curve $r=a\cos3\theta$ which is parallel to the initial line(horizontal axis).

here is my steps: $y=r\sin\theta=a\cos(3\theta)\sin\theta$

$dy/d\theta=a(\cos\theta\cos(3\theta)-3\sin(3\theta)\sin\theta)=0$.

I tried to use Product-to-Sum formula

->$\frac{1}{2}(\sin(π/2+2\theta)-\sin(π/2-4\theta))-\frac{3}{2}(\sin(π/2+2\theta)+\sin(π/2+4\theta))=0$

then I don't know how to continue..

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"Initial line" is the horizontal axis? –  J. M. Jul 12 '12 at 6:05
    
yes. Initial line means the x-axis. –  Vic. Jul 12 '12 at 6:43
    
A hint: you'll need to find the values of $\theta$ within $[0,\pi]$ such that $\cos\,2\theta=2\cos\,4\theta$... –  J. M. Jul 12 '12 at 6:46
    
I did it by using the identities for $\cos(3\theta)$ and $\sin(3\theta)$ in terms of $\cos\theta$ and $\sin\theta$ respectively. You should be able to derive them, or find them somewhere. By the way, it is $4\cos^3\theta-3\cos\theta$ and $3\sin\theta-4\sin^3\theta$. Alternately, you can use an identity for $\tan(3\theta)$ in terms of $\tan(\theta)$. There are neater ways, using product to sum identities. –  André Nicolas Jul 12 '12 at 6:47
    
I tried Product-to-sum identities but it makes the equation more complicated. I also tried to replace $\cos(3\theta)$ and $\sin(3\theta)$ with $4\cos^3\theta-3\cos\theta$ and $3\sin\theta-4\sin^3\theta$ and I get $4\cos^4\theta+12\sin^4\theta=3\cos^2\theta+9\sin^2\theta$. –  Vic. Jul 12 '12 at 7:14
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$\cos\theta\cos(3\theta)-3\sin(3\theta)\sin\theta=1/2(\cos2\theta+\cos4\theta-3(\cos2\theta-\cos4\theta))=0\implies \cos2\theta=2\cos4\theta$.$$$$ Now $\cos4\theta=2\cos^2 2\theta-1$.Putting this in equation and letting $\cos2\theta =t$ gives, $$t=2(2t^2-1)$$ $$\implies 4t^2-t-2=0$$ $$\implies t=\frac{1\pm \sqrt{33}}{8}$$ and then $\theta=\frac{\cos^{-1}(t)}{2}$ and then you can find the equation easily.

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