given a point P($\sin\theta$, $\cos \theta$), find the range of $\theta$ such that P lies in a given area

Question

A point P($\sin \theta, \cos \theta$), where $0\le\theta\le2\pi$. find the range of $\theta$ such that lies inside the triangle given by points $(0, 0)$, ($0, \sqrt[2]{\frac 32}$), ($\sqrt[2]{\frac 32}, 0$)

so, i have tried something.

now x coordinate is $\sin\theta$. thus, $0\le\sin\theta\le\sqrt[2]{\frac 32}$.

Similarly, $0\le\cos\theta\le\sqrt[2]{\frac 32}$.

then, $0\le\theta\le\sin^{-1}(\sqrt[2]{\frac 32})$ and $0\le\theta\le\cos^{-1}(\sqrt[2]{\frac 32})$

but, $cos^{-1}(\sqrt[2]{\frac 32})$ and $sin^{-1}(\sqrt[2]{\frac 32})$ and are imaginary.

$cos^{-1}(\sqrt[2]{\frac 32}) = 1.57079\iota$

why is this method incorrect??

Answer 1

Note that the point $P(\sin \theta , \cos \theta)$ lies on circle of radius $1$ centred $(0,0)$. Also the given area is a right triangle in first quadrant right angled at origin $(0,0)$.

Let the points be $A, B, C$ in the order you mention them. We need to find the point of intersection of line $BC$ and the circle $x^2 + y^2 = 1$.

The two points generated from these intersection will have each a parametric angle associated with it, let them be $\theta_1 , \theta_2$. The angle $\theta$ must lie in $(0, \theta_1) \cup (\theta_2, \frac{\pi}{2})$

Answer 2

Although it is true that $0\le \sin\theta, \cos\theta \le \sqrt{\frac32}$ for all points inside the triangle, there are also points outside of the triangle who satisfy this. In fact, instead of a triangle, you have found the bounds of $\theta$ for the square with its bottom-left corner at the origin and side lengths of $\sqrt{\frac32}$.

For the triangle, First we can implement what is easy: the bounds from the horizontal and vertical sides. In this case, they tell us that $\theta$ must satisfy $0\le\theta\le\frac\pi2.$ The last side is slightly more complicated, but it should be relatively straightforward.

For the last side, we must find its equation then proceed from there. In this case, that equation is $\sin\theta+\cos\theta=\sqrt{\frac32}.$ This should be intuitive. This can be simplified to $\sqrt2\cos\theta=\sqrt{\frac32}$, which is easily solved using algebra.

Using the line's equation instead of the rectangle's bounds you used will yield an answer of $\left[0, \frac\pi4-\arccos\left(\sqrt{\frac34}\right)\right]\cup\left[\frac\pi4+\arccos\left(\sqrt{\frac34}\right), \frac\pi2\right],$ which can be verified using a graphing calculator.

Answer 3

so this is my attempt, but please suggest some simpler method since it is difficult to solve like this without a scientific calc:

so as the value of $\theta$ differs, point P traverses a unit circle. to get the point inside the triangle, it is obvious that $0 \le\theta\le\frac\pi2$.

now we can find the intersecting points of circle with the hypotenuse of the triangle by solving the equations of the hypotenuse and the circle, i.e.,:

$x+y-\sqrt{\frac32} = 0$ and $x^2+y^2-1=0$. solving this we get

$x = $$\sqrt{3}\pm1\over2\sqrt{2}$ and $y=$$\sqrt{3}\mp1\over2\sqrt{2}$.

now taking slopes of lines made by joining the points to the origin $O(0,0)$, and using $\tan^{-1}$ to find the inclinations we get,:

$5\pi\over12$ and $\pi\over12$

thus the range of $\theta$ must be:

$[0, \frac\pi2] -[$$\pi\over 12$$, $$5\pi\over12$$]$

please do suggest some easier methods such that it can be done manually