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)=1-2y
\sqrt{y^2+1}\cos (x-t)=1-2y
\cos(x-t) =\dfrac{1-2y}{\sqrt{y^2+1}} , whose range is [-1,1]
0 \leq \dfrac{(1-2y)^2}{ y^2+1 } \leq 1
4y²-4y+1≤y²+1
3y²-4y≤y
3y(y-\dfrac{4}{3} )≤0
0≤y≤4/3
So the range of the function is [0,4/3]
Apply the double angle identities
\sin 2\alpha = \dfrac{2\tan \alpha }{1+\tan^2 \alpha }
\cos 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}
3y+yt^2 = t^2+2t+1
(1-y)t^2+ 2t+1-3y = 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(1-y)(1-3y)\geq 0
1-(1-3y-y+3y^2)\geq 0
3y^2-4y\leq 0
0 \leq y \leq \dfrac{4}{3}