A trigonometric equation - Finding the value of theta
I have to find the value of $\theta$ in this equation "$\cos2\theta \sin\theta = 1$"
The problem is that I have to solve using algebra (making equations...etc) like this one, for example: \begin{align} 4\sin^2\theta - 1 = 0\\ \sin^2\theta = 0.25\\ \sin\theta = +0.5, -0.5\\ \theta = 30, 150, 210, 330\end{align}
But for "$\cos2\theta \sin\theta = 1$", I figured, according to the fact that $\cos\alpha$ can be only between -1 and 1. And the same for $\sin\alpha$, that the only two expected solutions are when $\cos2\theta$ and $\sin\theta$ are both equal to 1 or -1 so that the product is equal to 1 as in the main equation. So $\theta = 270$. \begin{align} \cos(540)\sin(270) = 1\\ \cos(180)\sin(270) = 1\\ -1 * -1 = 1 \end{align}
I'm trying to find another way to solve it like the way I solved the other example $4\sin^2\theta - 1 = 0$..
$\endgroup$4 Answers
$\begingroup$You need formulae for double angle, e.g. in here: $ \cos (2 \theta) = 1 - 2 (\sin (\theta))^2 $
Then let $x = \sin (\theta)$ to get
$$ 1 = \cos (2 \theta) \sin (\theta) = (1 - 2 (\sin (\theta))^2) \sin (\theta) = (1-2x^2)x $$
Solving that third order equation gives as the only real root $x = -1$, hence $-1 = \sin (\theta)$ or
$$ \theta = \frac32 \pi $$
and of course you may add multiples of $2 \pi$.
$\endgroup$ 0 $\begingroup$The simplest solution is to put $\cos 2\theta = 1 - 2\sin^2\theta$, after which you get a cubic in terms of $\sin\theta$. Fortunately, that has an obvious solution using the Rational Root Theorem and you can easily factorise it.
$\endgroup$ 1 $\begingroup$Hint:
Use the identity-$\cos2\theta=\cos^2\theta-\sin^2\theta$
So,$$\cos2\theta\sin\theta=1$$
$$\implies(\cos^2\theta-\sin^2\theta)\sin\theta=1$$
Now,replace $\cos^2\theta$ by $1-\sin^2\theta$
So,$$(1-2\sin^2\theta)\sin\theta=1$$
$$\implies\sin\theta-2\sin^3\theta-1=0$$
Now,let,$\sin\theta=x$.So,this reduces to a cubic equation.Now,solve for $\sin\theta$ and just plug the value in the equation.
Hope this helps!!
$\endgroup$ 1 $\begingroup$Nice try and got $\sin \theta = -1$ and hence $\theta = 270^0$ as one root.
However, as other answerers have shown, the equation is reduced to a degree 3 polynomial (which is $2\sin^3\theta – \sin\theta + 1 = 0$), there should be 3 roots in total. Unless you can show the other 2 roots are imaginary, your work cannot be considered as complete.
Extra steps needed:-
Divide the polynomial by $(\sin \theta + 1)$ to get the other factor (which is $2 \sin^2 \theta – 2 \sin \theta + 1)$).
Show that the determinant of quotient factor is less than zero.
