Question
Find the range of the function y = \dfrac{\sin x+1}{\cos x -2}
Find the range of the function y = \dfrac{\sin x+1}{\cos x -2}
y = \dfrac{\sin x+1}{\cos x -2}
∵ \cos x-2 \ne 0
∴y\cos x-2y=\sin x+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]
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}
-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