# Coordinates of Points on a Circle

Alignments to Content Standards: G-GPE.B F-TF.C

Below is a picture of the unit circle in the plane with an angle of measure $a$: 1. What are the coordinates of $Q$ in terms of $a$? Explain.
2. For which value of $a$ between 0 and 360 does the sum of the coordinates of $Q$ take the largest value? What is this largest value?

## IM Commentary

The purpose of this task it to use geometry and algebra in order to understand the behavior of the trigonometric function $f(x) = \sin{x} + \cos{x}$. The task has been stated in an open ended fashion as there are natural solutions using geometry, or using the trigonometric identity $\sin{2x} = 2\sin{x}\cos{x}$, or algebraically solving a system of equations. Guidance is likely to be needed for any of the  approaches and some helpful suggestions might include:

• Students can be encouraged to make tables in order to conjecture the answer to part (b).  See https://www.illustrativemathematics.org/tasks/1868 for ideas on such a task that would lead nicely into this one.
• For the first solution, the tangent line to the unit circle at $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ can be given and students might also need encouragement to see that this line is precisely the set of points $(x,y)$ in the plane with $x + y = \sqrt{2}$ (the maximal value sought and achieved at the point where the tangent line touches the circle).
• For the second solution, a hint of squaring the expression $\sin{x} + \cos{x}$ may be sufficient.
• For the third solution, the base idea is similar to that of the first solution, namely studying the tangent line to the unit circle at $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$.  In this case, however, algebraic equations take the place of the intuitive geometric argument presented in the first solution to identify the largest value.

The theme of maximizing a function is a major component of mathematics coursework which lies beyond the purview of the Common Core.  While calculus is needed to maximize most functions, it is enlightening to find cases (e.g., quadratic functions, certain trig functions) where this process can be undertaken without calculus.  But if teachers are working with students in a calculus course, there is also a beautiful derivation of this result which uses the derivative. Differentiating the function $f(x) = \sin{x} + \cos{x}$ and then setting this equal to zero calculates the maximum (and minimum) values taken by $f$.  It is important for students to be exposed to problems like this where completely different viable approaches are available.  This helps to reveal a sense of unity in the standards and also encourages creative problem solving.

## Solutions

Solution: 1 Geometric Solution

1. The point $Q$ is on the unit circle and makes an angle of measure $a$ with the positive $x$-axis so we have  $Q = (\cos{a}, \sin{a})$.
2. We need to find a maximum value for the sum $x+y$ of the coordinates of $Q=(x,y)$ which, according to part (a), is $\cos{a} + \sin{a}$.   In order to get a sense for how big this quantity can be, we plot below several lines of the form $x + y = c$ for different numbers $c$ as indicated on the graph: We see that when $x + y = 1$ or $x + y = \frac{5}{4}$ these lines meet the circle in two points (the points whose coordinate sums are 1 and $\frac{5}{4}$ respectively). When $x + y = 2$ or $x + y = \frac{3}{2}$, the lines do not meet the circle at all.  For one value of $c$ between $\frac{5}{4}$ and $\frac{3}{2}$, these line meets the circle in exactly one point: in other words, for one value of $c$, this line is a tangent line to the circle. Below is a picture of the point $R = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ along with the tangent line to the circle at $R$. The equation of $\ell$ is $y = \sqrt{2} - x$. One way to see this is to observe that $m(\angle POR) = 45$ and so the equation of $\overleftrightarrow{OR}$ is $y = x$. So the slope of the tangent line to the unit circle at $R$ must be -1 since it is perpendicular to $\overleftrightarrow{OR}$. The tangent line goes through $R = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ and this allows us to find the equation $y = \sqrt{2} -x$. We could also write this equation as $x + y = \sqrt{2}$. Thus the blue line is the set of points $(x,y)$ whose coordinates add up to $\sqrt{2}$. The entire unit circle, except for the point $R$, is to the lower left hand side of this line. This means that all points $(x,y)$ on the unit circle, except for $R$, satisfy $x + y \lt \sqrt{2}$. So the maximum of the sum of the coordinates of points on the unit circle is $\sqrt{2}$ and this achieved only for the point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Solution: 2 Trigonometric solution to part (b)

We have $Q = (\cos{a},\sin{a})$ and so we are interested in finding the maximum value of $\cos{a} + \sin{a}$.  One way to do this is to look at the square of this quanitity:

\begin{align} (\cos{a} + \sin{a})^2 &= \cos^2{a} + 2\cos{a}\sin{a} + \sin^2{a}\\ &= 1 + \sin{2a}. \end{align}

Here we have used the trigonometric identities $\sin^2{a} + \cos^2{a} = 1$ and $2\sin{a}\cos{a} = \sin{2a}$. Now we need to maximize $1 + \sin{2a}$ but we can see that this happens when $\sin{2a} = 1$ or when $a = 45$ (since $a$ is in the first quadrant). So the maximum value of $(\cos{a} + \sin{a})^2$ is 2 and this means that the maximum value of $\cos{a} + \sin{a}$ is $\sqrt{2}$, achieved when $a = 45$.

Solution: Algebraic Solution

Since $\sin(45^\circ)=\cos(45^\circ)=\frac{\sqrt{2}}{2}$, we know that that the sum of the coordinates $x+y$ at that angle is $\sqrt{2}$.  We'll show that $x+y$ can't be any bigger than that by showing that if $c>\sqrt{2}$, there are no points on the unit circle with $x+y=c$.

To do this, we consider the system of equations

\begin{align*} x+y&=c\\ x^2+y^2&=1 \end{align*} Solving the first equation gives $y=c-x$. Substituting this into the second equation gives $x^2+(c-x)^2=1$, which simplifies to $2x^2-2cx+(c^2-1)=0$. We can use the quadratic formula to find the solutions to this equation, giving $$x=\frac{2c\pm\sqrt{(2c)^2-4(2)(c^2-1)}}{4}=\frac{c}{2}\pm \sqrt{8-4c^2}.$$ Now if $c>\sqrt{2}$, then $8-4c^2\lt 0$, which means there are no solutions to the system of equations. That is, the value of $c=\sqrt{2}$ is the largest possible value of $x+y$ for any point on the unit circle.