2
$\begingroup$

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..

  • 0
    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 $4\cos^4\theta+12\sin^4\theta=3\cos^2\theta+9\sin^2\theta$.2012-07-12

1 Answers 1

1

$\cos\theta\cos(3\theta)-3\sin(3\theta)\sin\theta=1/2(\cos2\theta+\cos4\theta-3(\cos2\theta-\cos4\theta))=0\implies \cos2\theta=2\cos4\theta$.$$ 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.