How to calculate the value of $\cos \theta$ if $\cos 3\theta$ is known

Question

If $\cos 3\theta = \frac{-11}{\sqrt{125}}$ then how can i calculate the the value $\cos \theta$ from there. I already tried so solve this by using $4\cos^3 \theta-3\cos \theta=\cos 3\theta $ but i was unable to solve the cubic equation.

Answer 1

Since you know complex numbers:

$$e^{3i\theta} = e^{(i\theta)^3} $$

which leads to:

$$ cos(3\theta) + i \cdot sin(3\theta) = (cos(\theta) + i \cdot sin(\theta))^3$$

Group the real and imaginary parts to get $cos(3\theta)$ and $sin(3\theta)$ as functions of powers of $sin(\theta)$ and $cos(\theta)$.

The equation you'll get for $sin(3\theta)$ will be a second-degree one. Use the identity $cos(x)^2 + sin(x)^2 = 1$ when necessary.

Answer 2

The tricky part here is to put the cubic equation into a form where factors can be identified. We could eliminate the radical by squaring the equation or by deriving an equation whose roots are squared, but these methods introduce extraneous roots.

The superior method is to render $\sqrt{125}=(\sqrt5)^3$ and with that, define $x=(\sqrt5)\cos\theta$. Thereby

$4(x/\sqrt5)^3-3(x/\sqrt5)+11/(\sqrt5)^3=0$

Now the $\sqrt5$ appears to odd powers in all terms allowing it to be cancelled. Clearing the remaining fractions then gives:

$4x^3-15x+11=0$

which can now be factored if the standard rational root search can give a root for $x$. Finding the root $x=1$ we obtain the factorization

$(x-1)(4x^2+4x-11)=0$

and the quadratic roots are found by the usual methods. Remember to render $\cos\theta=x/\sqrt5$ at the end.