How do I find the maximum and minimum values of $1−4\cos(2\theta)+3\sin(2\theta)$?

Question

To find the maximum of $$1 - 4\cos(2\theta) + 3\sin(2\theta) $$I tried: $$1-4(1)+3(1)=0.$$ To find the minimum I tried to substitute with the minimum values of sin and cos: $$1-4(-1)+3(-1)=2$$

I know I'm wrong, could someone explain why? And how I should go around obtaining the correct answer?

Thanks

Answer 1

Your objective is to maximize and minimize $y$ where $y = 1 - 4 \cos(2\theta) + 3\sin(2\theta)$, and find the one value of some one $\theta$ that will permit you to do that, not to evaluate by picking one $\theta$ at which the maximum/minimum value $\cos\theta$ is, a picking another $\theta = \alpha$ such that $\sin2\alpha$ is maximized or minimized, since then we have $2\theta_1 \neq 2\alpha = 2\theta_2$ appearing in the expression, where we must have the same $\theta$ to evaluate for $y$ AT THAT ONE $\theta$

Example:

Notice,

when $\theta = \pi/2$, $y = 1 - 4\cos(\pi) + 3\sin(\pi) = 1 - 4(-1) + 0 = 1 + 4 + 0 = 5$

When $\theta = 0, \;\;y = 1 - 4\cos(0) + 3\sin(0) = - 3$.

But we can do better than that...$y$ can be maximized a bit more, and we can do better on minimizing: $y$.. See if you can find a way to express $y$ as a function of either $\cos$ or $\sin$?

See arbautjc's trick in the comments above: when you have and expression which includes $a\cos(2\theta) + b\sin(2\theta)$ and can be expressed solely as a function of $\cos$: $$a\cos(2\theta) + b\sin(2\theta) = \sqrt{a^2+b^2} \cos (2\theta - \phi)$$ $$\text{ where}\;\;\cos \phi = \dfrac{a}{\sqrt{a^2 + b^2}}\;\text{ and}\;\;\sin \phi = \dfrac{b}{\sqrt{a^2 + b^2}}$$

Plotting the graph of $y$ will always provide you with some intuition, as well.

Answer 2

$$1-4 \cos (2 \theta)+3 \sin(2 \theta) = 1 - 5 \left({4 \over 5} \cos (2 \theta) - {3 \over 5} \sin (2 \theta) \right)$$ $$=1-5 \cos (2 \theta + \phi)$$ With $\cos \phi = {4 \over 5}$ and $\sin \phi = {3 \over 5}$. Now conclude.

Answer 3

Following is an algebraic way of finding the range of $

As $\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$

So, $1-4\cos2\theta+3\sin2\theta=\frac{(1+t^2)-4(1-t^2)+3(2t)}{1+t^2}=y$(say) putting $t=\tan\theta$

So, $t^2(y-5)-6t+y+3=0$

This is a Quadratic Equation in $t=\tan\theta$

As $\theta$ is real, so will be $t,$ hence the discriminant $(-6)^2-4(y-5)(y+3)\ge0$

$\implies y^2-2y-24\le0$

(i)$\implies (y-6)(y+4)\le 0\implies$ either $(y\le 6, y\ge -4\implies -4\le y\le 6)$ or $(y\le -4,y\ge 6)$ which is impossible

(ii)$\implies (y-1)^2\le 25=5^2\implies -5\le y-1\le 5\implies -4\le y\le 6$