﻿ Find the range of the function y = \dfrac{\sin x+1}{\cos x -2}

#### Question

Find the range of the function y = \dfrac{\sin x+1}{\cos x -2}

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

y = \dfrac{\sin x+1}{\cos x -2}

∵ \cos x-2 \ne 0

∴y\cos x-2y=\sin x+1

∴y\cos x-\sin x=2y+1
(1)

Let \cos t = \dfrac{y}{y^2+1}, \sin t = \dfrac{1}{y^2+1}

(1) is converted to

\sqrt{y^2+1}(\cos x\cos t-\sin x\sin t)=2y+1

\sqrt{y^2+1}\cos (x+t)=2y+1

\cos(x+t) =\dfrac{2y+1}{\sqrt{y^2+1}} , whose range is [-1,1]

0 \leq \dfrac{(2y+1)^2}{ y^2+1 } \leq 1

4y²+4y+1≤y²+1

3y²+4y≤y

3y(y+\dfrac{4}{3} )≤0

-4/3≤y≤0

So the range of the function is [-4/3,0]

Steven Zheng posted 2 years ago

Using the double angle identities

\sin 2\alpha = 2\sin \alpha \cos \alpha=\dfrac{ 2\sin \alpha \cos \alpha}{\sin^2 \alpha +\cos^2 \alpha } = \dfrac{2\tan \alpha }{1+\tan^2 \alpha }

\cos 2\alpha = \cos^2\alpha - \sin^2 \alpha = \dfrac{\cos^2\alpha - \sin^2 \alpha }{\cos^2 \alpha+\sin^2\alpha } =\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha }

Let t = \tan x

y = \dfrac{\sin x+1}{\cos x -2} = \dfrac{ \dfrac{2\tan \alpha }{1+\tan^2 \alpha }+1}{\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha } -2} =\dfrac{ \dfrac{2t }{1+t^2 }+1}{\dfrac{1-t^2 }{1+t^2 } -2}

= \dfrac{t^2+2t+1}{-1-3t^2}
(1)

-y-3yt^2 = t^2+2t+1

(3y+1)t^2+2t+1+y = 0

In order for the quadratic equation to have real roots, the discriminant

b^2-4ac should be larger than or equal to 0. So we have the following inequality.

2^2-4(3y+1)(1+y)\geq 0

1-(3y+3y^2+1+y)\geq 0

3y^2+4y\leq 0

-\dfrac{4}{3} \leq y \leq 0

Steven Zheng posted 2 years ago

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