Trigo Range

Geometry Level 3

Let x x and y y be the real numbers such that

x 2 + 2 cos y = 1 x^2+2\cos y=1

What is the range of x cos y x-\cos y ?

[ 1 , 3 + 1 ] \left[-1, \sqrt{3}+1 \right] [ 1 , 3 + 1 ] \left[1, \sqrt{3}+1 \right] [ 1 , 3 1 ] \left[-1, \sqrt{3}-1 \right] [ 1 , 3 1 ] \left[1, \sqrt{3}-1 \right]

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3 solutions

Chris Lewis
Nov 3, 2020

It's tempting to eliminate x x or y y using the condition, but this leads to a lot of case analysis. Instead, we can use Lagrangian multipliers: define f ( x , y , L ) = x cos y + L ( x 2 + 2 cos y 1 ) f(x,y,L)=x-\cos y + L \left(x^2+2\cos y-1 \right)

By setting the partial derivatives to zero, we can find the extreme values subject to the constraint. Differentiating: f x = 1 + 2 L x f y = ( 1 2 L ) sin y f L = x 2 + 2 cos y 1 \begin{aligned} f_x&=1+2Lx \\ f_y&=(1-2L)\sin y \\ f_L&=x^2+2\cos y-1 \end{aligned}

From f y = 0 f_y=0 , either sin y = 0 \sin y=0 or L = 1 2 L=\frac12 .


Case sin y = 0 \sin y=0 :

In this case, cos y = ± 1 \cos y=\pm1 . From f L = 0 f_L=0 , x 2 ± 2 1 = 0 x^2\pm2-1=0

For this to have a real solution, we need cos y = 1 \cos y=-1 ; hence x 2 3 = 0 x^2-3=0

and x = ± 3 x=\pm \sqrt3 .

When x = 3 x=\sqrt3 , x cos y = 1 + 3 x-\cos y=1+\sqrt3 . When x = 3 x=-\sqrt3 , x cos y = 1 3 x-\cos y=1-\sqrt3


Case L = 1 2 L=\frac12 :

From f x = 0 f_x=0 , we find x = 1 x=-1 .

Now from f L = 0 f_L=0 , we get cos y = 0 \cos y=0 ; so x cos y = 1 x-\cos y=-1 .


The largest of these values is 1 + 3 1+\sqrt3 ; the smallest is 1 -1 ; so the range is [ 1 , 1 + 3 ] \boxed{\left[-1,1+\sqrt3 \right]} .

It would be helpful if the expression x cos ( y ) x - \cos(y) was in terms of x x . By solving for cos ( y ) \cos(y) in the given equation we find that cos ( y ) = 1 x 2 2 \displaystyle \cos(y) = \frac{1 - x^2}{2} . Since cos ( y ) \cos(y) has a range of [ 1 , 1 ] [-1, 1] we must conclude that the domain of x x is [ 3 , 3 ] [-\sqrt3, \sqrt3] .

Substituting our expression for cos ( y ) \cos(y) in terms of x x into x cos ( y ) x - \cos(y) yields x 2 + 2 x 1 2 \displaystyle \frac{x^{2}+2x-1}{2} . This is a parabola with a minimum at ( 1 , 1 ) (-1, -1) and a maximum at ( 3 , 1 + 3 ) (\sqrt3, 1 + \sqrt3) because the domain of x x is bounded. The range of x cos ( y ) x - \cos(y) is therefore [ 1 , 3 + 1 ] \boxed{[-1, \sqrt3 + 1]} .

From x 2 + 2 cos y = 1 x^2 + 2\cos y = 1 , x = ± 1 2 cos y \implies x = \pm \sqrt{1-2\cos y} . Then x cos y = ± 1 2 cos y cos y x - \cos y = \pm \sqrt{1-2\cos y} - \cos y . Note that 1 2 cos y \sqrt{1-2\cos y} is real for π 3 y 5 π 3 \frac \pi 3 \le y \le \frac {5\pi}3 .

Consider the maximum value of x cos y = ± 1 2 cos y c o s y x - \cos y = \pm \sqrt{1-2\cos y} - cos y . Since both 1 2 cos y \sqrt{1-2\cos y} and ( cos y ) (-\cos y) are maximum when y = π y=\pi , then max ( x cos y ) = 1 2 cos π cos π = 3 + 1 \max(x-\cos y) = \sqrt{1-2\cos \pi} - \cos \pi = \sqrt 3 + 1 .

Now consider the minimum value. Since 1 2 cos y 0 \sqrt{1-2\cos y} \ge 0 , min ( x cos y ) \min(x-\cos y) occurs when x cos y = 1 2 cos y cos y x-\cos y = - \sqrt{1-2\cos y} - \cos y or when 1 2 cos y + cos y \sqrt{1-2\cos y} + \cos y is maximum. Let z = 1 2 cos y + cos y z = \sqrt{1-2\cos y} + \cos y . To find the maximum,

d z d y = sin y 1 2 cos y sin y Putting d z d y = 0 \begin{aligned} \dfrac {dz}{dy} & = \dfrac {\sin y}{\sqrt{1-2\cos y}} - \sin y & \small \blue{\text{Putting }\frac {dz}{dy}=0} \end{aligned}

sin y ( 1 1 2 cos y 1 ) = 0 { sin y = 0 y = π z = 3 1 1 1 2 cos y 1 = 0 y = π 2 z = 1 \sin y \left(\dfrac 1{\sqrt{1-2\cos y}}-1\right) = 0 \implies \begin{cases} \sin y = 0 & \implies y = \pi & \implies z = \sqrt 3 - 1 \\ \dfrac 1{\sqrt{1-2\cos y}}-1 = 0 & \implies y = \frac \pi 2 & \implies z = 1\end{cases}

Therefore min ( x cos y ) = 1 \min(x-\cos y) = - 1 and the range of x 2 cos y x - 2\cos y is [ 1 , 3 + 1 ] \boxed{\left[-1, \sqrt 3+1\right]} .

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