Question
Find the maximum value of the function y = \dfrac{\sin x\cos x}{1+\sin x+\cos x}
Find the maximum value of the function y = \dfrac{\sin x\cos x}{1+\sin x+\cos x}
(1+\sin x+\cos x)(1-\sin x-\cos x)
=1-(\sin x+\cos x)^2
=-2\sin x\cos x
Therefore,
y = \dfrac{\sin x\cos x}{1+\sin x+\cos x}
=\sin x+\cos x-1
=\sqrt{2} \cdotp (\dfrac{\sqrt{2} }{2}\sin x+ \dfrac{\sqrt{2} }{2}\cos x)-1
=\sqrt{2}(\sin x \cos 45\degree +\cos x\sin 45\degree )-1
=\sqrt{2}\sin (x-45\degree )-1
So the maximum value of the function is \sqrt{2}-1