Question

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

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer 1

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=1 -2y
(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]

Steven Zheng posted 1 month ago


Answer 2

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}

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

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}

Steven Zheng posted 1 month ago

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